ELEMENTS 


DIFFERENTIAL  AND  INTEGRAL 
CALCULUS, 


EXAMPLES  AND   APPLICATIONS. 


BY 

JAMES   M.  TAYLOR, 

PROFESSOR   OF   MATHEMATICS,   MADISON  UNIVERSITY. 


)>K° 


BOSTON: 

PUBLISHED  BY  GL\N,  HEATH,  &  CO. 
1884. 


I 


Entered  according  to  Act  of  Congress,  in  the  year  1884,  by 

JAMES  M.  TAYLOR, 
in  the  Office  of  the  Librarian  of  Congress,  at  Washington. 


J.  8.  Cushing  &  Co.,  Printers,  115  High  Street,  Boston. 


PEEFAOE 


THE  object  of  the  following  treatise  is  to  present  sim- 
ply and  concisely  the  fundamental  problems  of  the 
Calculus,  their  solution,  and  more  common  applications. 

Since  variables  are  its  characteristic  quantities,  the 
first  fundamental  problem  of  the  Calculus  is,  To  find  the 
ratio  of  the  rates  of  change  of  related  variables.  To  ena- 
ble the  learner  most  clearly  to  comprehend  this  problem, 
the  author  has  employed  the  conception  of  rates,  which 
affords  finite  differentials  and  the  simplest  demonstration 
of  many  principles.  The  problem  of  Differentiation  hav- 
ing been  clearly  presented,  a  general  method  of  its  solu- 
tion is  obtained  by  the  use  of  limits.  This  order  of 
development  avoids  the  use  of  the  indeterminate  form  -, 
and  secures  all  the  advantages  of  the  differential  nota- 
tion. Many  principles  are  proved,  both  by  the  method 
of  rates  and  that  of  limits,  and  thus  each  is  made  to 
throw  light  upon  the  other. 

In  a  final  chapter,  the  method  of  infinitesimals  is  briefly 
presented;  its  underlying  principles  having  been  previ- 
ously established. 

The  chapter  on  Differentiation  is  followed  by  one  on 
Integration ;  and  in  each,  as  throughout  the  work,  there 


IV  PREFACE. 

are  numerous  practical  problems  in  Geometry  and  Me- 
chanics, which  serve  to  exhibit  the  power  and  use  of 
the  science,  and  to  excite  and  keep  alive  the  interest 
of  the  student. 

In  writing  this  treatise,  the  works  of  the  best  Ameri- 
can, English,  and  French  authors  have  been  consulted ; 
and  from  these  sources  the  most  of  the  examples  and 
problems  have  been  obtained. 

The  author  is  indebted  to  Professors  J.  E.  Oliver,  and 
J.  McMahon  of  Cornell  University,  and  Professor  O. 
Root,  Jr.,  of  Hamilton  College,  for  valuable  suggestions; 
and  to  Messrs.  J.  S.  Cushing  &  Co.  for  the  typograph- 
ical excellence  of  the  book. 

J.  M.  TAYLOR. 

Hamilton,  N.Y., 
Nov.,  1884. 


CONTENTS. 


CHAPTER  I.  h 

INTRODUCTION. 

Section.  Page. 

1.  Definition  of  variable  and  constant 1 

2.  Definition  of  function  and  independent  variable 1 

3.  Classification  of  functions 2 

4.  Definition  of  continuous  variable  and  continuous  function 2 

5.  Definition  of  the  limit  of  a  variable 3 

6.  Limits  of  equal  variables 3 

7.  Limit  of  the  product  of  a  constant  and  a  variable 4 

8.  Limit  of  the  product  of  two  or  more  variables 4 

9.  Limit  of  the  quotient  of  two  variables 4 

10.  Limit  of  the  sum  of  two  or  more  variables 4 

11.  Definition  of  uniform  change 5 

12.  Definition  of  increment 5 

13.  Definition  of  differential 5 

14.  Illustrations  of  differentials 6 

15.  Definition  of  iyiclination,  slope,  and  tangent 7 

16.  Geometric  signification  of  -£ 7 

dx 

17.  Limit  of  the  ratio  of  the  increments  of  y  and  x 8 


CHAPTER  n.)( 

DIFFERENTIATION. 

18.  Definition  of  differentiation.     Differentiation  of  ax2 10 

Algebraic  Functions. 

19.  Differential  of  the  product  of  a  constant  and  variable 10 

20.  Differential  of  a  constant 11 

21.  Differential  of  the  sum  of  two  or  more  variables 11 


v> 


VI  CONTENTS. 

Section.  Page. 

22.  Differential  of  the  product  of  two  variables 12 

23.  Differential  of  the  product  of  several  variables 13 

24.  Differential  of  a  fraction • 14 

25.  Differential  of  a  variable  with  a  constant  exponent 14 

2G.    General  symbol  for  the  differential  oif[x).     Examples    ....  15 

27.  Definition  of  an  increasing  and  a  decreasing  function 17 

28.  Definition  of  derivative 17 

29.  Measure  of  rate  of  change 18 

30.  Signification  of  ^ 18 

dt 

31.  Signification  of  f'(x)  or  clJl 18 

dx 

32.  Limit  of  the  ratio  of  Ay  to  Ax.     Applications 19 

33.  Definition  of  velocity  and  acceleration.     Examples 23 

Logarithmic  and  Exponential  Functions. 

34.  Differential  of  a  logarithmic  function 24 

35.  The  greater  the  base,  the  smaller  the  modulus .  25 

36.  Naperian  system 25 

37.  Differential  of  c* 26 

38.  Differential  of  if 26 

39.  Logarithmic  differentiation.     Examples 26 

Trigonometric  Functions. 

40.  Definition  of  the  unit  of  angular  measure 29 

41.  Differential  of  sin  x  and  cos  x 29 

42.  Differential  of  tana; 30 

43.  Differential  of  cot  x 30 

44.  Differential  of  sec  x 30 

45.  Differential  of  cosec  x .    .  31 

46.  Differential  of  vers  x 31 

47.  Differential  of  covers  x 31 

48.  Limit  of  the  ratio  of  an  arc  to  its  chord 31 

49.  Differentiation  of  sin  x  by  the  method  of  limits.     Examples    .    .  32 

Anti-Trigonometric  Functions. 

50.  Differential  of  sin_1x 35 

51.  Differential  of  cos_1x 35 

52.  Differential  of  tan~x:r 35 

53.  Differential  of  cot"1* 36 

54.  Differential  of  sec_1x 36 


CONTENTS.  Vll 

Section.  Page. 

55.  Differential  of  cosec_1x 30 

56.  Differential  of  vers"1^ ' 36 

57.  Differential  of  covers"1^.     Examples 36 

Miscellaneous  examples 39 


CHAPTER  III. 

INTEGRATION. 

58.  Definition  of  integral  and  integration.     Sign  of  integration    ...  43 

59.  Elementary  principles 43 

60.  Fundamental  formulas      44 

61.  Statement  of  formulas  1  and  2.     Examples 46 

62.  Auxiliary  formulas.     Examples 50 

63.  Trigonometric  differentials.     Examples 55 

64.  Definite  integrals.     Examples 58 

Applications  to  Geometry  and  Mechanics. 

65.  Rectification  of  curves.     Examples 60 

66.  Areas  of  plane  curves.     Examples 61 

67.  Graphical  representation  of  any  integral 63 

68.  Areas  of  surfaces  of  revolution.     Examples 63 

69.  Volumes  of  solids  of  revolution.     Examples 65 

70.  Fundamental  formulas  of  mechanics.     Examples Q6 

1.  Formulas  for  uniformly  accelerated  motion. 

2.  Motion  down  an  inclined  plane. 

3.  Motion  down  a  chord  of  a  vertical  circle. 

4.  Values  of  v  and  s  when  a  varies  directly  as  t. 

5.  Geometrical  representation  of  the  time,  velocity,   distance, 

and  acceleration. 

6.  Path  of  a  projectile. 

7.  Path,  velocity,  and  acceleration  of  a  hody  whose  velocity 

in  each  of  two  directions  is  given. 

CHAPTER  TV. 

SUCCESSIVE    DIFFERENTIATION. 

71.  Successive  derivatives 71 

72.  Signification  of f"lx),f'"(x),fn(x).    Examples 71 

73.  Successive  differentials 73 

74.  Relations    between    successive    differentials    and    derivatives. 

Examples 73 


Vlll  CONTENTS. 

CHAPTER  V. 

SUCCESSIVE    INTEGRATION    AND    APPLICATIONS. 

Section.  Page. 

75.  Successive  integration.     Examples 76 

76.  Problems  in  mechanics 77 

1.  Time  and  velocity  of  a  falling  body  below  the  surface  of 

the  earth. 

2.  Maximum  velocity  with  which  a  falling  body  can  reach  the 

earth. 

3.  Velocity  of  a  body  falling  from  the  sun. 

4.  Velocity  of  a  body  falling  in  the  air. 

5.  Velocity  of  a  body  projected  into  a  medium. 

6.  Velocity  of  a  body  sliding  down  a  curve. 

7.  The  cycloidal  pendulum  isochronal. 

8.  The  length  and  equation  of  the  catenary. 

CHAPTER  VI. 

INDETERMINATE   FORMS. 

77.  Value  of  functions  assuming  an  indeterminate  form.     Examples      86 

Evaluation  by  Differentiation. 

78.  The  form  -.     Examples 87 

0 

79.  The  form  ■—■     Examples 88 

80.  The  forms  0  •  go  and  cc  —  co.     Examples 89 

81.  Functions  whose  logarithms  assume  the  form  ±0  •  co.     Examples  91 

82.  Compound  indeterminate  forms.     Examples 91 

83.  Evaluation  of  derivatives  of  implicit  functions.     Examples     .    .  92 

CHAPTER  VH. 

DEVELOPMENT   OF   FUNCTIONS   IN   SERIES. 

84.  Definition  of  series,  convergent  infinite  series,  and  sum  of  infinite 

series 94 

85.  Definition  of  the  development  of  a  function 94 

86.  Definition  of  Taylor's  formula 95 

87.  Proof  of  Taylor's  formula 95 

88.  Proof  of  Maclaurin's  formula 96 


CONTENTS.  ix 

Section.  Page. 

89.  Proof  of  the  binomial  theorem 97 

90.  Development  of  log„(.r  +  >j) 97 

91.  Development  of  a*+J 98 

92.  Development  of  (a  +  x)m 98 

93.  Development  of  sin  a: 98 

94.  Development  of  cos  x 98 

95.  Exponential  series 99 

9G.  Logarithmic  series 99 

97.  Failure  of  Taylor's  and  Maclaurin's  formulas 102 

98.  Lemma 102 

99.  Completion  of  Taylor's  and  Maclaurin's  formulas 103 

100.  A  second  complete  form 104 

101.  Proof  that  ??  =  0  as  n  =  00 105 

[n 

102.  Maclaurin's  formula  develops  ax 105 

103.  Maclaurin's  formula  develops  sin  x  and  cos  x 100 

104.  The  logarithmic  series  holds  for  x >— 1  and  <+l 106 

105.  Development  of  (l  +  x)m  holds  for  x  >— 1  and  <+l 107 

106.  One  form  of  the  development  of  (a  +  x)m  always  holds  ....  108 

107.  Development  of  tan_1x,  and  value  of  tt 109 

108.  Development  of  sin"1:*:,  and  value  of  it 109 

109.  Geometric  proof  that  f[a  +  h)  =f(a)  +  hf'(a  +  eh).    Examples  110 


CHAPTER  VIII. 

MAXIMA    AND    MINIMA. 

110.  Definition  of  maximum  and  minimum 112 

111.  The  critical  values  of  x 112 

112.  Two  methods  of  examining/(:r)  at  critical  values  of  x   .   .    .    .  113 

113.  Maxima  and  minima  occur  alternately 114 

114.  Principles  facilitating  the  solution  of  problems 114 

Examples  and  geometric  problems 115 

CHAPTER  IX. 

FUNCTIONS  OF  TWO  OR  MORE  VARIABLES,  AND  CHANGE  OF  THE 
INDEPENDENT  VARIABLE. 

115.  Applicability  of  previous  rules  for  differentiation 125 

116.  Definition  of  partial  differential 125 

117.  Definition  of  total  differential 125 


X  CONTENTS. 

Section.  Pcge. 

118.  Definition  of  partial  derivative 125 

119.  Definition  of  total  derivative 125 

120.  The  value  of  the  total  differential 126 

121.  The  signification  of  partial  derivatives.     Examples 126 

122.  One  method  of  finding  the  total  derivative 127 

123.  Formulas  for  finding  the  total  derivative.     Examples     ....  128 

124.  Implicit  functions 130 

125.  Formula  for  the  derivative  of  an  implicit  function.     Examples  130 

126.  Successive  derivatives  of  an  implicit  function.     Examples    .    .  131 

127.  Successive  partial  differentials  and  derivatives 132 

128.  The  order  of  differentiations  is  indifferent.     Examples  ....  133 

129.  Formulas  for  successive  differentials 134 

130.  Change  of  the  independent  variable 135 

131.  Forms  of  successive  derivatives,  dx  being  variable.     Examples  135 

CHAPTER  X.J| 

TANGENTS,  NORMALS,  AND  ASYMPTOTES. 

132.  The  equation  of  a  tangent 139 

133.  The  equation  of  a  normal.     Examples 139 

134.  Length  of  tangent,  normal,  subtangent,  and  subnormal.  Exam- 

ples      141 

135.  Important  principle  in  the  method  of  limits 143 

136.  Length  of  polar  subtangent,  subnormal,  etc 143 

137.  A  second  proof.     Examples 145 

138.  Rectilinear  asymptotes.     Examples 146 

139.  Asymptotes  determined  by  inspection  or  expansion.   Examples  149 

140.  Asymptotes  to  polar  curves.     Examples 151 


CHAPTER  XI. 

DIRECTION   OF   CURVATURE,    SINGULAR    POINTS,    AND    CURVE 
TRACING. 

141.  Direction  of  curvature.     Examples      154 

142.  Definition  of  singular  points 156 

143.  Points  of  inflexion.     Examples 156 

144.  Points  of  inflexion  on  polar  curves 157 

145.  Definition  of  multiple  point      157 

146.  At  a  multiple  point  -JL  has  two  or  more  values 158 

dx 


CONTENTS.  XI 

Section.                                                 .                                               ,.  Page. 

147.  At  a  multiple  point  L!L  assumes  the  form  - 158 

dx  0 

148.  Examination  of  a  curve  for  multiple  points.     Examples    .    .    .  159 
140.    Shooting  points  and  stop  points 103 

150.  Curve  tracing.     Examples     ...» • 103 

151.  Tracing  polar  curves.     Examples 108 


CHAPTER  XII. 

CURVATURE,  EVOLUTES,  ENVELOPES,  AND  ORDER  OF  CONTACT. 

152.  Direction  of  a  curve 170 

153.  Definition  of  curvature 170 

154.  Measure  of  the  curvature  of  a  circle 170 

155.  Formula  for  curvature 171 

150.    Radius  of  curvature 171 

157.  Radius  of  curvature  in  polar  curves 172 

158.  Intersection  of  a  curve  and  its  circle  of  curvature 172 

159.  Exception  to  §  158 172 

AGO'.   Definition  of  an  evolute 174 

^101.   Deduction  of  the  equation  of  an  evolute.     Examples 174 

102.  A  normal  to  an  involute  is  a  tangent  to  its  evolute 177 

103.  Length  of  an  arc  of  an  evolute 173 

104.  Tracing  an  involute  from  its  evolute 179 

105.  Definition  of  a  variable  parameter 180 

100.    Definition  of  an  envelope 180 

107.  Deduction  of  the  equation  of  an  envelope      180 

108.  An  envelope  is  tangent  to  each  curve  of  the  series.     Examples  181 

109.  Definition  of  contact  of  different  orders      183 

170.  Intersection  of  curves  at  their  point  of  contact 183 

171.  Osculating  curves 184 

Examples 185 

CHAPTER   XIII. 

INTEGRATION   OF    RATIONAL    FRACTIONS. 

172.  Decomposition  of  rational  fractions 188 

173.  Simple  factors  of  the  denominator  real  and  unequal.     Exam- 

ples    188 

174.  Some  of  the  simple  factors  real  and  equal.     Examples  ....  190 

175.  Some  imaginary  and  unequal     Examples 191 

170.    Some  imaginary  and  equal.     Examples 193 


Xll  CONTENTS. 


CHAPTER  XIV. 

INTEGRATION    BY   RATIONALIZATION. 

Section.  Page. 

177.  Rationalization  of  a  differential 195 

c 

178.  Differentials  containing  surds  of  the  form  xTi.     Examples     .    .  195 

c 

179.  Surds  of  the  form  (a  +  bx)a.     Examples 196 

180.  Surds  of  the  form  Va  +  bx  +  x2 196 

181.  Surds  of  the  form  Va  +  bx  —  xl 197 

182.  Binomial  differentials 199 

183.  Conditions  of  rationalization  of  xm(a  -f  bxn)ldx.     Examples     .  199 

CHAPTER   XV. 

INTEGRATION    BY   PARTS    AND   BY   SERIE8. 

184.  Formula  for  integration  by  parts.     Examples 203 

185.  Formulas  of  reduction.     Examples 204 

186.  Integration  of  <p( x) (log x)ndx.     Examples 210 

187.  Integration  of  x"amxdx.     Examples 211 

188.  Integration  of  sinm:r  cosnxdx.     Examples 211 

189.  Integration  of  xn  sin  ax  dx  and  xn  cos  ax  dx.     Examples     .    .    .  213 

190.  Integration  of  e0*  sin":r  dx  and  eax  cos"x  dx.    Examples.    .    .    .  213 

191.  Integration  of — 215 

a  +  b  cos  x 

192.  Integration  of  f(x)  sin" lxdx,  etc.     Examples 215 

193.  Integration  by  series.     Examples 216 

194.  Integration  often  leads  to  higher  functions 217 


CHAPTER  XVI. 

LENGTH  AND  AREAS  OF  PLANE  CURVES,  AREAS  OF  SURFACES  OF 
REVOLUTION,  VOLUMES  OF  SOLIDS. 

195.  Examples  in  rectification  of  plane  curves 218 

196.  Rectification  of  polar  curves.     Examples 219 

197.  Examples  in  quadrature  of  plane  curves 220 

198.  Quadrature  of  polar  curves.     Examples 222 

199.  Examples  in  quadrature  of  surfaces  of  revolution 223 

200.  Examples  in  cubature  of  solids  of  revolution 224 

201.  Equations  of  curves  deduced  by  aid  of  the  Calculus.     Exam- 

ples     225 


CONTENTS.  Xiii 
CHAPTER   XVII. 

THE   METHOD    OF    INFINITESIMALS. 

Section.  Page. 

202.  Definition  of  infinitesimals  and  infinites 22G 

203.  Orders  of  products  and  quotients 227 

204.  Geometric  illustration  of  infinitesimals  of  different  orders    .    .  227 

205.  First  fundamental  principle  of  infinitesimals 228 

206.  Rule  for  differentiation 229 

207.  Second  fundamental  principle  of  infinitesimals 231 

208.  Integration  as  a  summation 232 

209.  Definition  of  centre  of  gravity 233 

210.  Centre  of  gravity  of  any  plane  surface 233 

211.  Centre  of  gravity  of  any  plane  curve 234 

212.  Centre  of  gravity  of  any  solid  of  revolution 234 

Examples 235 


ELEMENTS  OF  THE  CALCULUS. 


Elements  of  the  Calculus. 


chapter  I. 


INTRODUCTION. 


1.  In  the  Calculus  there  are  two  kinds  of  quantities  considered, 
variables  and  constants. 

A  Variable  is  a  quantity  that  is,  or  is  conceived  to  be,  con- 
tinually changing  in  value.  Variables  are  usually  represented 
by  the  final  letters  of  the  alphabet. 

A  Constant  is  a  quantity  whose  ATalue  is  fixed  or  invariable. 
Constants  are  usually  represented  by  figures  or  the  first  letters 
of  the  alphabet.     Particular  values  of  variables  are  constants. 

In  the  Calculus  the  locus  of  an  equation  is  conceived  as  traced 
by  a  moving  point  called  the  Generatrix.  If  a  =  ob,  the  locus 
of  ar+?/2=a2  is  the  circle  abcd.  Now, 
as  the  generatrix  traces  this  circle,  its 
coordinates,  x  and  ?/,  continually  change 
in  value,  and  are  therefore  variables ; 
while  a  retains  the  value  ob,  and  is 
therefore  a  constant. 

2.  Functions  and  Independent  Vari- 
ables. One  variable  is  a  function  of  an- 
other, when  the  two  are  so  related  that 

any  change  of  value  in  the  second  produces  a  change  of  value 
in  the  first. 

For  example,  the  area  of  a  varying  square  is  a  function  of 
its  side  ;  the  volume  of  a  variable  sphere  is  a  function  of  its 
radius  ;  all  mathematical  expressions  depending  on  x  for  their 
values,  as  ax*,  foe4  +  car,  sin  a;,  log  a,  etc.,  are  functions  of  a;. 

An  independent  variable  is  one  to  which  any  arbitrary  value 
or  law  of  change  may  be  assigned  ;  as,  x  in  or5,  x  in  sin  X',  etc. 


2  INTRODUCTION. 

The  symbol  f(x)  is  used  to  denote  any  function  of  x,  and  is 
read  "  function  of  as."  When  several  functions  of  x  occur  in 
the  same  investigation,  we  employ  other  sjTnbols,  as  f'(x), 
F(x),  <f>(x),  etc.,  which  are  read  "/  prime  function  of  x," 
"i'' function  of  x,"  "  <£  function  of  aj,"  etc.  According  to  this 
notation,  y  =f(x)  represents  any  equation  between  x  and  y, 
when  solved  for  y. 

3.  Algebraic  and  Transcendental  Functions. — An  algebraic 
function  is  one  that  is  expressed  in  terms  of  its  variable  or 
variables,  by  means  of  algebraic  signs,  without  the  use  of 
variable  exponents;  as,  ax5  — 2 ex2,  5x?  —  x,  etc. 

All  functions  not  algebraic  are  called  transcendental.  These 
are  sub-divided  into  exponential,  logarithmic,  trigonometric,  and 
anti-trigonometric. 

An  Exponential  function  is  one  in  which  the  variable  enters 
the  exponent ;  as,  a",  y**. 

A  Logarithmic  function  is  one  that  involves  the  logarithm  of 
a  variable  ;  as,  log  x,  log  (fix  +  c). 

The  sine,  cosine,  tangent,  etc.,  of  a  variable  angle  are  called 
Trigonometric  functions. 

The  symbol  sin-1a;,  read  "anti-sine  of  x,"  denotes  the  angle 
whose  sine  is  x.  Sin-1a;,  cos_1x,  tan-1#,  etc.,  are  called  Inverse 
Trigonometric,  or  Anti-Trigonometric,  functions. 

4.  A  variable  is  Continuous,  or  varies  continuously,  when,  in 
passing  from  one  value  to  another,  it  passes  successively  through 
all  intermediate  values. 

A  Continuous  function  is  one  that  is  constantly  real,  and 
varies  continuously,  when  its  variable  varies  continuously. 
Some  functions  are  continuous  for  all  real  values  of  their  vari- 
ables, others  only  for  those  between  certain  limits.  Thus,  if 
y  =  ax  +  b,  or  y  =  sin  x,  y  is  evidently  a  continuous  function  of 
x  for  all  real  values  of  x ;  but,  if  y  =  ±  Vr*  —  x2,  y  is  continuous 
only  for  values  of  x  between  the  limits  —  r  and  +  r. 

The  Calculus  treats  of  variables  and  functions  only  between 
their  limits  of  continuitj' ;  hence  all  the  values  of  x  and  f(x) 
that  it  considers  are  represented  geometrically  by  the  coordi- 
nates of  the  points  of  the  plane  curve  whose  equation  fo  yz=f(x). 


THEORY   OF   LIMITS.  3 

Theory  of  Limits. 

5.  For  convenience  of  reference,  we  give  here  a  brief  state- 
ment of  the  theory  of  limits. 

The  Limit*  of  a  variable  is  a  constant  quantity  which  the 
variable,  in  accordance  with  its  law  of  change,  approaches 
indefinitely  near,  but  which  it  never  reaches.  The  variable  may 
be  less  or  greater  than  its  limit. 

Thus,  if  the  number  of  sides  of  a  regular  polygon  inscribed 
in  or  circumscribed  about  a  circle  be  indefinitely  increased, 
the  area  of  the  circle  will  be  the  limit  of  the  area  of  either 
polygon,  and  the  circumference  will  be  the  limit  of  the  peri- 
meter of  either.  When  the  polygons  are  inscribed,  the  variable 
area  and  perimeter  are  less  than  their  limits  ;  and,  when  the  poly- 
gons are  circumscribed,  the  variable  area  and  perimeter  are 
greater  than  their  limits. 

By  increasing  the  number  of  terms,  the  sum  of  the  series, 
l+£+4+i+  etc. ,  can  be  made  to  approach  2  as  nearly  as  we 
please,  but  it  cannot  reach  2  ;  hence  2  is  the  limit  of  the  sum. 

Again,   if   a   point   starting  from   a   move  the   distance  AC 
(=^ab)   the  first  second,  the  distance 
cd  (=^-cb')  the  second  second,  and  so       ' ' ' ' 

v  -  '  A  C  D  B 

on,  ab  will  evidently  be  the  limit  of  the  K    2 

line  traced  by  this  point. 

Cor.  The  difference  between  a  variable  and  its  limit  is  a  vari- 
able whose  limit  is  zero. 

6.  If  tico  variables  are  continually  equal,  and  each  approaches 
a  limit,  their  limits  are  equal;  that  is,  if  x  =  y,  and  limit  (x) 
=  a,  and  limit  (y)  =  b,  a  =  b. 

For,  since  x  =  y,  a  —  x  =  a  —  y;  hence,  as  a  is  the  limit  of  «,  it 
is  also  of  y  (§  5,  Cor.).  Since  a  and  b  each  is  a  limit  of  y,  and 
y  cannot  approach  two  unequal  limits  at  the  same  time,  a  =  b. 

Cor.  If  one  of  two  continually  equal  variables  approaches  a 
limit,  the  other  approaches  the  same  limit. 

*  The  student  should  carefully  note  the  two  senses  in  which  the  word 
limit  is  used.  In  the  theory  of  limits,  a  limit  is  a  value  which  the  variable 
cannot  reach ;  in  other  cases,  as  in  §  4,  a  limit  is  the  greatest  or  the  least 
value  which  the  variable  actually  reaches. 


4  INTRODUCTION. 

7.  The  limit  of  the  product  of  a  constant  and  a  variable  is  the 
product  of  the  constant  and  the  limit  of  the  variable;  that  is,  if 
limit  (x)  =  a,  limit  (ex)  =  ca. 

Let  v  =  a  —  x ; 

then  cx=  ca  —  cv. 

Now  limit  (cv)  =  0,  since  limit  (v)  =  0  ; 

hence  limit  (ex)  =  limit  (ca  —  cv)  =  ca. 

8.  The  limit  of  the  variable  product  of  two  or  more  variables 
is  the  product  of  their  limits;  that  is,  if  limit  (x)  =  a,  and  limit 
(y)  =  b,  limit  (xy)  =  ab. 

Let  v  =  a  —  x,  and  v1  =  b  —  y; 

then  x=  a  —  v,  and  y  =  b  —  v^ ; 

.•.  xy  =  ab  —  (ai\-{-  bv  —  vv{) . 

Now         limit  (m?!+  bv  —  vv1)*=  0  ; 
hence  limit  (xy)  =  limit  [a&  —  (ai\-\-  bv  —  vv1)']=  ab. 

In  like  manner,  the  theorem  is  proved  for  n  variables. 

9.  The  limit  of  the  variable  quotient  of  two  variables  is  the 
quotient  of  their  limits;  that  is,  if  limit  (x)  =  a,  and  limit  (y)  =b, 
limit  (x  -=-  y)  =  (a  -5-  b)  .f 

Let  z  =  x-^-y, 

and  c  =  limit  (z) ,  or  limit  (x  -f-  y) . 

Then       x  =  yz;  .•.  a  =  bc;  §§6,8. 

.-.  limit  (a;-5-y)[=c3  =  a-s-&. 

10.  The  limit  of  the  variable  sum  of  a  finite  number  of  vari- 
ables is  the  sum  of  their  limits;  that  is,  if  limit  (x)  =  a,  limit 
(y)  =  b,  limit  (z)  =  c,  etc., 

limit  (x+y-4-z-| )  =  a-fb  +  cH 

Let  v  =  a  —  x,  vx  =  b  —  y,  v2  =  c  —  z,  etc. 

Then        %  +  y  +  z-\-  •'• 

=  (a +  &+.-C +  ».)—(«  +  **■+«*+—) ; 

.*.  limit  (x  +  y  -\-z  4-  •••) 

=  limit  [_(a  +  b  +  c+  '••)-(v  +  vl  +  v.2+  ---y] 
—  a  +  b  4-c  4-  ••• 

*  When  v  and  v1  have  unlike  signs,  the  difference,  ai\  +  bv  —  vvv  may  be- 
come zero  for  particular  values  of  v  and  vv  but  it  cannot  remain  zero,  since 
xy  is  variable.      The  same  is  true  of  the  difference,  v  +  v1  +  v2  •] — ,  in  §  10. 

t  This  principle  does  not  hold  when  the  limit  of  the  divisor  is  zero. 


INCREMENTS    AND   DIFFERENTIALS.  0 

Con.  When  the  product,  quotient,  or  sum  of  two  or  more  vari- 
ables is  equal  to  a  constant,  the  iiroduct,  quotient,  or  sum  of  their 
limits  is  equal  to  the  same  constant. 

11.  The  Change  of  a  variable  is  Uniform,  when  its  value 
changes  equal  amounts  in  equal  arbitrary  portious  of  time.  In 
all  other  cases   the   change    is  variable. 

Thus,  if  from  a  toward  b  a  point  move      } — j — £    c    ^ — — ^ 
equal  distances,  as  a«,  ab,  be,  etc.,  in  equal  Fi    3 

arbitrary  portions  of  time,  the  increase  of 

the  line  traced  will  be  uniform.  Again,  if  the  motion  of  a  point 
along  a  straight  line  be  uniform,  the  change  of  each  of  its 
rectilinear  coordinates  will  evidently  be  uniform. 

12.  An  Increment  of  a  function  or  variable  is  the  amount  of 
its  increase  or  decrease  in  any  interval  of  time,  and  is  found  by 
subtracting  its  value  at  the  beginning  of  the  interval  from  its 
value  at  the  end.  Hence,  if  a  variable  is  increasing,  its  incre- 
ment is  positive ;  and,  if  it  is  decreasing,  its  increment  is 
negative.  An  increment  of  a  variable  is  denoted  by  writing 
the  letter  A  before  it;  thus,  Ax,  read  "increment  of  x,"  is  the 
symbol  for  an  increment  of  x.  If  y  =f(x) ,  Ax  and  Ay  repre- 
sent corresponding  increments,  that  is,  the  increments  of  x  and 
y  in  the  same  interval  of  time. 

Let  am  be  the  locus  of  y  =f(x)  referred  to  the  rectangular 
axes  ox  and  oy.     If,  when  x  =  oa,         y  d 

Ax  =  on  —  oa  =  ab  ;    then 
Ay  =bp'  —  ap  =  ep'  ;  if,  when  x  =  oc, 
Ax  =  cf  ;  then  Ay  =  fii  —  cd  =  —  nd. 

In  the  last  case  Ay  is  negative,  but 
it   is    properly    called   an   increment,  Flg-  4- 

since  it  is  what  must  be  added  to  the  first  value  to  produce 
the  second. 

13.  The  Differential  of  a  function  or  variable  at  any  value  is 
what  ivoidd  be  its  increment  in  any  interval  of  time,  if  at  that 
value  its  change  became  uniform.     Hence,  the  differential  of  a 


V 

£ 

H 

H 

r 

6 


INTRODUCTION. 


variable  is  positive  or  negative,  according  as  the  variable  is 
increasing  or  decreasing.  The  interval  of  time,  though  arbitrary, 
must  be  the  same  for  a  function  as  for  its  variable. 

If  the  change  of  a  variable  be  uniform,  any  actual  increment 
may  evidently  be  taken  as  its  differential. 

The  differential  of  a  variable  is  represented  by  writing  the 
letter  d  before  it;  thus,  dx,  read  "differential  as,"  is  the  symbol 
for  the  differential  of  x.  When  the  symbol  of  a  function  is  not 
a  single  letter,  parentheses  are  used;  thus,  d(x*)  and  d(a?  —  2x) 
denote  the  differentials  of  x"  and  ar  —  2x. 


i-'k 


14.  Illustrations  of  Differentials.    Conceive  a  variable  right 

triangle  as  generated  by  the  perpendicular  moving  uniformly  to 

the  right.      Let  y  represent  its  area,  x  its 

base,  and  2 ax  its  altitude;  then  y=ax2.    Let 

bh  be  Ax  estimated  from  the  value  ab(=  x')  , 

then  bhmc  will  be  Ay.     But,  if  the  increase 

of  the  area  became  uniform  at  the  value  abc,  „k^ 

a 

the  increment  of  the  area  in  the  same  time 
would  evidently  be  bhoc  ;  hence,  bhoc  and 
bh  may  be  taken  as  the  differentials  of  y  and 
x,  when  x=  x'.  But  bhoc  =  2ax'dx,  hence, 
in  general,  dy\_  =  d{ax2)~\=2axdx.  If  a=l,  y=a2,  sm<\dy=2xdx. 
Here  Ay  =  dy-j-  triangle  com. 

The  signification  of  dy=2ax'dx  is  evidently  that,  when 
x  =  x',  y,  the  area,  is  changing  in  units  of  surface  2  ax'  times  as 
fast  as  x  is  in  linear  units. 

Again,  let  opw  be  the  locus  of  y=f{x),  referred  to  the  axes 
ox  and  oy.  Conceive  the  area  between  ox  and 
the  curve  as  traced  b}'  the  ordinate  of  the  curve 
moving  uniformly  to  the  right.  Let  z  repre- 
sent this  area,  and  let  ab  be  Ax  estimated 
from  the  value  oa  (=&');  then  abp'p  =  Az. 
But,  if  the  increase  of  z  became  uniform  at  the 
value  oap,  its  increment  in  the  same  interval 
would  evidently  be  abdp  ;  hence  ab  and  abdp  may  be  taken  as 
the  differentials  of  .x  and  z  respectively,  when  x  =  x'. 


GEOMETRIC    ILLUSTRATIONS   OF  DIFFERENTIALS.  7 

Hence  dz  =  abdp=apc?x-  =  y'dx  ;  or,  in  general,  dz  =  ydx, 
which  evidently  means  that  z  is  changing  y  times  as  fast  as  x. 

Area  above  the  axis  of  x  being  positive,  area 'below  it  is 
negative ;  hence,  where  the  curve  lies  below  the  axis  of  3J,  the 
area  decreases  as  x  increases,  and  ydx  is  negative  as  it  should  be. 

Here  Az  =  dz  +  area  pdp'. 

15.  The  Inclination  of  a  straight  line  referred  to  rectangular 
axes  is  the  angle  included  between  the  axis  of  abscissas  and  the 
line.  The  direction  of  a  line  with  respect  to  the  axis  of  x  is 
determined  by  its  inclination. 

The  Slope  of  a  line  is  the  tangent  of  its  inclination.  Thus,  in 
Fig.  7,  hzp  is  the  inclination  of  za,  and  tanHZP  is  the  slope  of  z.v. 

The  direction  of  motion  of  the  generatrix  of  a  straight  line  is 
constant,  while  the  direction  of  motion  of  the  generatrix  of  a 
curve  is  variable. 

A  Tangent  to  a  curve  at  any  point  is  the  straight  line  that 
passes  through  that  point,  and  has  the  same  direction  as  the 
curve  at  that  point ;  or,  a  tangent  to  a  curve  at  any  point  is  the 
straight  line  that  the  generatrix  would  trace,  if  its  direction  of 
motion  became  constant  at  that  point.  The  slope  of  a  curve  at 
any  point  is  the  slope  of  its  tangent  at  tl.at 
point.  Thus,  if,  in  Fig.  7,  pa  is  a  tangent  to 
the  curve  at  p,  tannzA  is  the  slope  of  the 
curve  at  p. 

16.  Geometric  Signification  of  -X    Let  mn 

be  the  locus  of  y=f{x),  and  let  x'  be  the 
abscissa  of  any  point  upon  it,  as  p.  If  at  P 
the  motion  of  the  generatrix  of  the  curve  Fig.  7. 

became  uniform  along  the  tangent  pa,  it  is 
evident  that  the  change  of  each  of  its  coordinates  would  also 
become  uniform.  Hence  pe,  ea,  and  pa  may  be  taken  respec- 
tively as  the  differentials  of  x,  y,  and  the  length  of  the  curve, 
when  aj  =  x' ;  for  they  are  what  would  be  the  simultaneous 
increments  of  these  variables,  if  the  change  of  each  became 

uniform  at  the  value  considered.     Therefore  —  =  —  =  tan  epa 

dx      pe 


o'z 


INTKODUCTION. 


=  tannzA,   which  is  the  slope  of  the  curve  at  p.     Hence,  in 

general,  -^  is  the  slope  of  the  curve  y  =  f  (x)  at  any  point  (x,  y). 
clx 

Cor.  1.  If  ea,  or  cly,  be  c  times  as  great  as  pe,  or  dx,  y  is 
evidently  increasing  c  times  as  fast  as  x,  when  x  =  01. 

Cor.  2.  If  s  represent  the  length  of  the  curve  mn,  PA  =  ds-, 
and  els2  =  clx2  +  dy2,  in  which  ds2  denotes  the  square  of  ds. 

17.   Limit  of  the  Ratio  of  the  Increments  of  y  and  x. 

Let  mn  be  the  locus  of  y  =/(#),  and  ed,  a  tangent  at  p,  any 

point  upon  it ;  then  the  slope  of 


Let 


this  tangent  =  %  (§  1G), 
dx 

mn  [=  pc]  =  Ax,  when  estimated 

from  the  value  om,  then  cp'=A?/. 

Draw   the    secant   pp'  ;    then 

Aw 

— -  =  the  slope  of  the  secant  pp'. 

Ax 

Conceive  Ax  to  approach  0  as 
its  limit ;   then  the  slope  of  the 
secant  will  approach  the  slope  of  the  tangent  as  its  limit.* 


Fig.  8. 


.     limit 
'  '  Ax  =  0 


Ay 

Ax 


clx 


Hence,  the  ratio  of  the  differential  of  a  function  to  that  of  its 
variable  is  the  limit  of  the  ratio  of  their  increments,  as  these  incre- 
ments approach  zero  as  their  limit. 


Cor.  1. 


It  is  evident  that  ^L,  or  limi* 
dx        a*  =  0 


Ay 

Ax 


,  is  finite,  except 


where  the  locus  of  y=f(x)  is  parallel  or  perpendicular  to  the 
axis  of  x,  where  it  is  0  or  <x. 

*  This  statement,  if  not  sufficiently  evident,  may  be  demonstrated  as 
follows :  "When  the  arc  pap'  is  continuous  in  curvature,  and  this  arc  can 
always  be  made  so  small  that  it  will  be  continuous,  the  slope  of  the  secant 
pp'  is  equal  to  the  slope  of  a  tangent  to  the  arc  pap'  at  some  point,  as  a. 
Now,  as  the  arc  pup'  approaches  zero  as  its  limit,  the  point  a  approaches  p 
as  its  limiting  position ;  hence  the  slope  of  the  secant  pp'  approaches  the 
slope  of  the  tangent  pd  as  its  limit. 


limit 

T  Ax  =  0 


is  read  "  the  limit  of  — ■-  as  Ax  approaches  0  as  its  limit.: 
Ax 


LIMIT   OF   THE   IlATIO   OF   INCREMENTS. 


9 


Cor.  2.   If  —  be  constant,  the  locus  of  y=f(x)  is  evidently 

Ax  A         7 

a  straight  line,  in  which  case  — -  =  -f- 

°  Ax      dx 

Con.  3.    A  tangent  to  mn  at  r  is  evidently  the  limiting  posi- 
tion of  the  secant  pp'  as  p'  approaches  p  and  arc  p'p  =  0. 

The  following  is  another  proof  of  the  important  principle 
established  above :  — 

Second   Proof.*      Conceive  the  area  between  ox  and  the 
curve  OFn  (Fig.  9)  as  traced  by  the  ordinate 
of   the  curve  moving   to   the   right.      Let  z  T 
represent  this  area,  and  let  ab  be  Ax  esti- 
mated from  the  value  oa  ;  then  Ay  =  dp',  and 

AZ  =  ABp'p. 

o 


NOW  ABDP  <  ABP'P  <  ABP'JI  ; 

.-.  y'Ax  <Az<  (y'-j-  Ay)  Ax.  Fig.  9. 

Dividing  by  Ax,  we  have 

y  <^,<y+*y- 

Az 
Whence  —  differs  from  y'  less  than  y'-\-  Ay  does  ;  but 

Aa; 


But 
Hence 


limit 
Ax 

dz  _ 
dx~* 

limit 
Ar  =  0 


limit 


To[2/'+A^  =  2/''  '-Si 


'Az' 
Ax 


=  y 


§  H. 


"Az" 
Aa; 


dz 
dx' 


*  This  demonstration  assumes  that  any  function  of  x  may  be  repre- 
sented graphically  by  the  area  between  a  curve  and  the  axis  of  x.  That 
many  functions  of  x  may  be  thus  represented  is  very  evident,  and  that  any 
may  be  follows  from  §  67. 


CHAPTER   II. 
DIFFERENTIATION. 

18.  Differentiation  is  the  operation  of  finding  the  differential 
of  a  function.  The  sign  of  differentiation  is  d  ;  thus  d  in  d(xi) 
indicates  the  operation  of  differentiating  ar3,  while  the  whole 
expression  cK^)  denotes  the  differential  of  x?  (see  §  13). 

To  differentiate  ax2,  let  y  =  ax2,  and  let  x'  and  y'  be  any  cor- 
responding values  of  x  and  y  ;  then 

y'=ax'2.  (1) 

Let  Ax  be  any  increment  of  x  estimated  from  the  value  x\ 
and  Ay  the  corresponding  increment  of  y  ;  then 

y'+  Ay  =  a  (x'-j-  Ax)2  =  ax'2  +  2  ax' Ax  +  (Ax)2.         (2) 

Subtracting  (1)  from  (2),  we  have 

Ay  =  2  ax' Ax  +  (Aa;)2,  or  =^  =  2ax'+  Ax.  (3) 


limit  [~Ay 
*x  =  °\_Ax_ 


A^it0l.2aX'+AX^  §6' 


■  =  2  ax',  or,  in  general,  dy  =  2axdx.  §  17. 

dx 

By  this  general  method  we  could  differentiate  any  other  func- 
tion, but  in  practice  it  is  more  expedient  to  use  the  rules  which 
we  proceed  to  establish. 


Algebraic  Functions. 

19.  The  differential  of 'the  'product  of  a  constant  and  a  vari- 
able is  the  product  of  the  constant  and  the  differential  of  the 
variable. 


ALGEBRAIC   FUNCTIONS. 


11 


We  are  to  prove  that  d  (ay)  =  ady,  in  which  y  is  some  func- 
tion of  x.  Let  u  =  ay,  and  let  x'  represent  any  value  of  x,  and 
y'  and  u'  the  corresponding  values  of  y  and  u  ;  then 

u'=ay'.  (1) 

Let  Ax  represent  any  increment  of  x,  estimated  from  the  value 
x',  and  let  Ay  and  Au  represent  the  corresponding  increments 
of  y  and  u ;  then 

u'+  Aw  =  a  (?/'+  Ay)  =  ay'+  aAy.  (2) 

Subtracting  (1)  from  (2),  member  from  member,  we  have 

Aw  =  aAy. 


Aw        A?/ 
—  a— -■ 


Ax 

limit 
A*  =  0 


Ax 

Au 
Ax 

dy 


limit 
Ax  =  0 


A/ 
Ax 


limit 


Ax 


du  _ 
dx        dx 


§§  6,  7. 
§  17. 


Hence,  as  x'  is  any  value  of  if,  we  have  in  general,  by  multi- 
plying both  members  by  dx, 

du  [=  d  (aw)]  =  ady. 


,'?)=d(i/)=^. 

,a/         \a    J       a 


20.    27*e  differential  of  a  constant  is  zero. 
This  is  evident,  since  the  increment  of  a  constant  in  any 
interval  of  time  is  zero. 


21.  The  differential  of  a  x>olynomial  is  the  algebraic  sum  of 
the  differentials  of  its  several  terms. 

We  are  to  prove  that  d  {v+y— z+a)  =  dv+dy— dz,  in  which 
v,  y,  and  z  are  functions  of  x. 

Let  %i  =  v-\-y —  z  +  a,  and  let  x'  represent  any  value  of  », 
and  i'\  y',  z',  and  u'  the  corresponding  values  of  v,  y,  z,  and  u ; 
then 


v'-\-y'—  z'  +  a. 


(1) 


12 


DIFFERENTIATION. 


Let  Ax  represent  any  increment  of  x,  estimated  from  the  value 
re',  and  Av,  Ay,  Az,  and  Aw  the  corresponding  increments  of  w, 
y,  z,  and  w  ;  then 


«'  + Am=  v'+ Aw  +  w'+ Aw  — (z'+Az)  +  a. 
Subtracting  (1)  from  (2)  we  have 
Aw  =  Aw  +  Aw  —  Az. 


(2) 


Aw  _  Aw  ,  Aw  _  Az 
Ax      Ax      Ax      Ax 


limit 
A*  =  0 


A?/" 
Are 


.  dw  _  dy      dw 
da;      dec     dx 


limit 
Ax  =  0 

_dz_ 
dx 


"Aw  I  Aw  _  Az~] 
Ace      Ace      Arc  J 


§6. 
§§  10,  17. 


Hence,  as  x'  is  any  value  of  re,  we  have  in  general 
dw  [=  d  (w  +  y  —  z  +  «)]  =  c^  +  dy  —  dz. 

22.  T7ie  differential  of  the  product  of  two  variables  is  the  first 
into  the  differential  of  the  second,  plus  the  second  into  the  differen- 
tial of  the  first. 

We  are  to  prove  that  d  (wz)  =  ydz  +  zdy,  in  which  w  and  z  are 
functions  of  x. 

Let  w  =  wz,  and  let  re'  represent  any  value  of  x,  and  w',  z',  and 
w'  the  corresponding  values  of  w,  z,  and  w  ;  then 

u'=y'z'.  (1) 

Let  Arc  represent  anj-  increment  of  re  estimated  from  the  value 
re',  and  Aw,  Az,  and  Aw  the  corresponding  increments  of  w,  z, 
aud  it ;  then 

w'+  Aw  =  (w'+  Ay)  (z'+  Az) 

=  ?/'z'+?/'Az  +  z'Aw  +  AzAw.  (2) 

Subtracting  (1)  from  (2)  we  have 
Aw  =  w'Az  +  z'Aw  -f-  AzAw. 


^  =  yAz  +  (zf  +  A2)Ay> 
Arc       .    Ax  Ax 


ALGEBRAIC    FUNCTIONS. 


18 


limit 
Ax=0 


Am" 

Ax 


limit 


y  — 

Ax 


+ 


limit 
Ax  =  0 


<*^>il} 


du       ,dz   ,    ,dy 

—  =  y' \-z  '-£• 

dx         dx        dx 


§§  7,  8. 


B 

Fig.  10. 


Hence,  as  x'  is  any  value  of  x,  we  have  in  general 
dxi  [_=d  (yz)  ]  =  ydz  +  zdy . 

To  obtain  this  result  geometrically,  let  z  and  y  represent  the 
variable  altitude  and  base  of   a   rectangle 
conceived  as  generated  b}'  the  side  z  moving     "  go 

to  the  right,  and  the  upper  base  y  moving 
upward  ;  then  zy  =  its  area. 

If,  at  the  value  dcba  (Fig.  10),  dz  =  ah, 
and  dy  =  ce,  d(area)  =  cefb  +  bgha  ;  since 
cefb  +  bgha  is  evidently  what  would  be  the 
increment  of  the  area  of  the  rectangle  in 
the  assumed  interval,  if  at  the  value  dcba  the  increase  of  its 
area  became  uniform. 

Hence,  d(zy)  =  d(area)  =  cefb  +  bgha  =  zdy  +  ydz. 

Here      A  (zy)  =  d  (zy)  +  bgof. 

23.  The  differenticd  of  the  jwoduct  of  any  number  of  variables 
is  the  sum  of  the  products  of  the  differential  of  each  into  cdl  the 
rest. 

We  are  to  prove  that  d(xyz)  =  yzdx  -\-xzdy  -\-xydz,  in  which 
y  and  z  are  functions  of  x. 

Let  u  =  xy,  then  d(xyz)  =  d(uz) . 

But        d(uz)  =  zdu  +  udz  §  22. 

=  zd(xy)  +  xydz 

=  yzdx  +  xzdy  -f  xydz. 
.'.  d(xyz)  =  yzdx  -f  xzdy  +  xydz. 

In  a  similar  manner,  the  theorem  may  be  demonstrated  for 
any  number  of  variables. 


14  DIFFERENTIATION. 

24.  The  differential  of  a  fraction  is  the  denominator  into 
the  differential  of  the  numerator,  minus  the  numerator  into  the 
differential  of  the  denominator,  divided  by  the  square  of  the 
denominator. 


z  are 


We  are  to  prove  that  d(  ±  j  =  —    ~   ,  in  which  y  and 

functions  of  x. 

V 
Let  u  =  -i  then  uz  =  y. 

z  a 

.\udz-\-zdu  =  dy. 

cty  —  -dz 
dy  —  udz  z  zdy  —  ydz 

z  z  z2 

Cor.    a  (  -    = - = -,  since  da  =  0  ;  that  is,  the 

\xj  or  x- 

differential  of  a  fraction  with  a  constant  numerator  is  minus  the 

numerator  into  the   differential  of  the  denominator  divided  by 

the  square  of  the  denominator. 


25.  The  differential  of  a  variable  affected  zuith  any  constant 
exponent  is  the  product  of  the  exponent,  the  variable  with  its  expo- 
nent diminished  by  one,  and  the  differential  of  the  variable. 

I.  When  the  exponent  is  a  positive  integer. 

If  n  is  a  positive  integer,  xn  =  x  •  x  •  x  to  n  factors  ;  hence 
we  have 

d(xn)  =  d  (x  •  x  •  x  to  n  factors) 

=  xn~xdx  +  xn~Hx  +  etc.  to  n  terms  §  23. 

=  nxn~idx. 

II.  Wlien  the  exponent  is  a  positive  fraction. 

m 

Let  y  =  x«,  then  yn  =  xm.  (1) 

Differentiating  (1),  we  obtain 
nyn~xdy  =  mx^'^dx. 


ALGEBRAIC    FUNCTIONS.  15 


7        m  x""1  ,        m  af-1?/  7        ra  xm~lxn 
.-.  dv  = r  ax  = -  ax  = dx 

n  ?/  n     yn  n      xm 

—  i 
=  —xn     dx. 
n 

III.  When  the  exponent  is  negative. 

Let  y  =  x~n,  n  being  integral  or  fractional ;  then 

y=\-  (1) 

xn 
Differentiating  (1),  we  have 

dy  =  —  ?=£—  dx  =  -  nx-n~Hx.  §  24,  Cor. 

or" 

For  a  proof  of  this  theorem,  which  includes  the  case  of  in- 
commensurable exponents,  see  §  39,  Ex.  25. 

Assuming  the  binomial  theorem,  let  the  student  prove  this 
rule  by  the  general  method  of  differentiation. 

dx 


Cor.  d(Va)  =  \x~~*dx  = 


2Va 


26.    The  general  symbol  for  the  differential  of /(a;)  is/'  (x)  dx  ; 
hence,  if  y  =f(x),  dy  =f'(x)dx. 

Examples. 
Differentiate 

1.  ar3  +  8x  +  2ar.  Ans.    (3x* +  8 +4x)  dx. 

d  (ar3  +  8  x  +  2 x2)  =  d  (a?)  +  d  (8 a>)  +  d  (2 x2) ,  §  21 . 

d(a7,)  =  3x2dx,  §  25. 

d(8a.-)  =  8d.r,  §  19. 

d(2x2)  =  4xdx,  §§  19,  25. 
.'.  d  (ar5  +  8x  +  2a,-2)  =  (3ar  +  8  +  4 x)  dx. 

2.  ?/=  3  oar  —  5nx  —  8  m.  dy=  (6«a  —  5n)dx. 

dy  =  d  (3ax2—onx—8m)  =  d(3axr)  —d(5nx)  —  d(8m). 
[The  differentials  of  equals  are  equal,  and  §  21.] 


16  DIFFERENTIATION. 

3.  f(x)  -  5  ax2  -  3  Z^x3  -  abx*. 

f  (x)  dx  =  ( 1 0  ax  -  9  b2x?  -  4  aba?)  dx. 

4.  f(x)  =  a5  +  5  ft2.*,-3  +  7  a^x5. 

f'(x)  dx  =  (15  b-x?  +  35  a¥)  dx. 

5.  y  =  ax*  +  for*  +  c.  d?/  = tr- dx. 

2y/x 

6.  y2  =  2px.  dy  _jj< 

d(?/2)  =  d(2px).  dx~  y 

b2v 

7.  a2y2  +  Zrx2  =  a2&2.  d?/  = —  dx. 

a*/ 

8.  f(x)  =  (o  +  ax2)*.  f'(x)dx  =  -f (&  +  aar)*aaxfcc. 

9.  7/  =  (l+2x-0)(l+4x3).  ety  =  4x(l  +  3a?  +  lOx^dx. 

dy  =  (1  +  2x2)d(l  +  4x3)  +  (1  +  4x3)d(l  +  2x2) . 

.  n  x  +  a2  ,  &  —  a2     7 

10.  v  =  — ' dy  = dx. 

3      x  +  b  y      {x  +  b)2 

=  (X  +  b)d(x  +  a2)  -  (x  +  a2)d(x  +  b) 
{x  +  b)* 

11.  /(x)  = — ^— =•  f'(x)dx  =  — 4aa? .    dx. 
JK  '      b-2x2  J  K  J  (b-2x2)2 

a  —  Sx 


12.    y  =  (a  -f-  x)  Va  —  x.  dy  =  —  dx. 

2Va  — as 


2x4  -,.  N  7        8a2xi-4x5 

-s ;■  /  (#)  dx  = — -  ( 

a3 -a?  v  '  {ar-x2)2 


14.  /(x)=    -*-f.  /'(x)dx=      m*"        dx. 


i-xy  v  (i  —  x) 


15.    /(x)  =  -vl^t^-  /'(X)  = l  ; 

16«  /(*)=„    "^  /'(*)  = 


17.    ?/  = 

VI +X2 


(1-x2)*  "   v  '       (1-x2)! 

x 


DERIVATIVES.  17 

18.  2 xy2  —  ay2  =  Xs.  dy  =  — ■ — — — *—  clx. 

4  xy  —  2  ay 

19.  f(x)  = -•  f'(x)dx  = — J——  eta. 


20.    y 


4^  +  3^  (4.r  +  .r-)- 

ar5 


a2  —  x2 


21.  /(a-)=Vaa;  +  Vc2iC3.  /'(a,-)  =  ^ +  3c* 

2V.r 

22.    -• dx. 

(If  +  x2)*  (b2  +  x')< 

yV  ;      (1+x)2  J  V  ;      (1  +  z)3 

24.  /(s)=    2^Jli.  /'(*)  = 


25.    (»-2?/)(6-3.r)  =  (c-x2)(l-?/). 

dy  _  b  —  4  a*  -f-  6  y  —  2  xy 
dx       xr  —  6x  —  c-\-2b 

27.  An  Increasing  function  is  one  that  increases  when  its 
variable  increases ;  hence  it  decreases  when  its  variable  de- 
creases. 

A  Decreasing  function  is  one  that  decreases  when  its  variable 
increases ;    hence    it    increases   when    its    variable   decreases. 

Thus,  ax  and  a"  are  increasing,  and  -  and  a  —  x  are  decreasing 

x 
functions  of  x. 

28.  Derivatives.  The  ratio  of  the  differential  of  a  function  to 
the  differential  of  its  variable  is  called  the  derivative  of  the 
function.     This  ratio  is  called  also  the  differential  coefficient  and 

the  derived  function.    If  y  =/(as),  dy  =zf(x)  dx,  or  -j-  =f(x)  ; 
hence  —  and/* (x)  are  general  symbols  for  the  derivatives  of  y 

and /(a?).    If  f(x)  =  a,-6, /' (x)  =  6a^  ;  if  y=a?,  &[»/(»)]  =  3aj» 

dx 


18  DIFFERENTIATION. 

29.  The  Measure  of  the  rate  of  change  of  a  variable  at  a 
given  instant  is  what  would  be  its  increment  in  a  unit  of  time, 
if  at  that  instant  its  change  became  uniform.  This  measure 
of  rate  is  generally  called  the  rate.  Hence,  the  rate  will  be 
positive  or  negative,  according  as  the  variable  is  increasing 
or  decreasing.  Thus,  when  we  say  that  the  distance  of  a  train 
from  the  station  was  changing,  at  a  given  instant,  at  the  rate 
of  +30  miles  an  hour,  we  mean  that  this  distance  ivouhl  have 
increased  thirty  miles  in  an  hour,  if  at  that  instant  its  increase 
had  become  uniform. 

If  the  change  of  a  variable  is  uniform,  the  actual  increment 
of  the  variable  in  a  unit  of  time  is  the  measure  of  its  rate. 

30.  Signification  of  — .   Let  t  represent  time  ;  then,  any  vari- 

dt 

able,  as  y,  is  evidently  some  function  of  t.  Since  time  changes 
uniformly,  dt  may  represent  any  increment  or  interval  of  time. 
If  dt  equals  the  unit  of  time,  then  by  definition  dy  equals  the 
measure  of  the  rate  of  change  of  y  ;  and,  if  dt  is  n  times  the  unit 
of  time,  dy  is  n  times  the  rate  of  change  of  y  ;  hence,  whatever 

be  the  value  of  dt,  ^  is  the  rate  of  change  ofy. 
dt 

31.  Signification  of  f(x),  or  %    If  y=f{x),  ^L=f\x)^:, 

ax  dt  dt 

that  is,  f'(x),  or  ^,  is  the  ratio  of  the  rate  of  change  of  f(x) ,  or 
y,  to  that  of  x. 

Hence,  the  derivative  of  a  function  exjwesses  the  ratio  of  the  rate 
of  change  of  the  function  to  that  of  its  variable;  and  a  function 
is  an  increasing  function  or  a  decreasing  one,  according  as  its 
derivative  is  positive  or  negative. 

Cor.  The  same  function  of  x  may  be  an  increasing  function 
for  some  values  of  x,  and  a  decreasing  one  for  other  values. 

Thus,  since -,  the  derivative  of  —.  is  -f  when  x  <  0,    and 

x3  xr 

—  when  x  >  0,  — ,  is    an    increasing  function  when  a;<0,  and 

ar 
a  decreasing  function  when  x  >  0. 


APPLICATIONS.  19 

32.  When  the  change  of  y  is  uniform,  it  is  evident  that  — 

is  the  rate  of  change  of  y.     When  the  change  of  y  is  variable, 

the  value  of  — ^  evidently  lies  between  the  greatest  and  the  hast 

At 
values  of  the  rate  of  change  of  y  during  the  time  At ;  hence,  the 

smaller  At  is  taken,  the  nearer  -^  approaches  the  rate  of  change 

At 
of  y  at  the  beginning  of  At. 

'    is  the  rate  of  change  of  y  at  the  beginning 


Hence     lim.U 
of  At. 

Hence,   limita 


At 

— ^    is  the  ratio  of  the  rate  of  change  of  y  to 


that  of  x.      Thus,   without   the   aid  of  a   locus,  we  see   that 
=  ^,  since  each  equals   the  ratio  of   the   rate   of 


limit 
Ax  =  0 


'Ay 
Ax 


dx 


change  of  y  to  that  of  x. 

Applications. 

1.  The  area  of  a  circular  plate  of  metal  expanded  by  heat 
increases  how  man}'  times  as  fast  as  its  radius  ?  If,  when  the 
radius  is  two  inches,  it  is  increasing  at  the  rate  of  .01  inch  a 
second,  how  fast  is  the  area  increasing  at  the  same  time  ? 

Let  x  =  the  radius,  and  y  the  area  of  the  plate  ;  then  y  =  ttx2, 
.-.  dy  =  2ttX(1x  ;  that  is,  the  area  is  increasing  in  square  inches 
2ttx  times  as  fast  as  the  radius  is  in  linear  inches.  Let  dt  =  a 
second  ;  then,  when  x  =  2,  dx  =  .01  inch  ;  .-.  dy=2irxdx—  .047r 
sq.  in.  ;  that  is,  the  area  is  increasing  .04 7r  sq.  in.  a  second  at 
the  instant  considered. 

2.  The  volume  of  a  spherical  soap-bubble  increases  how  many 
times  as  fast  as  its  radius?  When  its  radius  is  3  in.,  and  is 
increasing  at  the  rate  of  2  in.  a  second,  how  fast  is  its  volume 
increasing  ? 

Here  dy  =  4  -n-xrdx  ;  that  is,  the  volume  is  increasing  in  cubic 
inches  4-ar  times  as  fast  as  the  radius  is  in  linear  inches.  The 
volume  is  increasing  72 tt  cu.  in.  a  second  at  the  instant  con- 
sidered. 


20  DIFFERENTIATION. 

3.  A  boy  is  running  on  a  horizontal  plane  in  a  straight  line 
towards  the  base  of  a  tower  50  metres  in  height.  He  is 
approaching  the  top  how  many  times  as  fast  as  he  is  the  foot  of 
the  tower?  How  fast  is  he  approaching  the  top,  when  he  is  500 
metres  from  the  foot,  and  running  at  the  rate  of  200  metres  a 
minute? 

Let  x  and  y  respectively  represent  in  metres  the  distances 
of  the  boy  from  the  foot  and  the  top  of  the  tower ;  then 
y2  =  x-  -f-  (50)2,  etc.  Ans.    199  metres  a  minute. 

4.  A  light  is  4  metres  above  and  directly  over  a  straight 
horizontal  side-walk,  on  which  a  man  If  metres  in  height  is 
walking  away  from  the  light.  The  farthest  point  of  the  man's 
shadow  is  moving  how  many  times  as  fast  as  he  is  walking? 
The  man's  shadow  is  lengthening  how  many  times  as  fast  as  he  is 
walking?     How  fast  is  the  shadow  lengthening,  and  its  farthest 

point  moving,  when  the  man  is  walking 
at  the  rate  of  50  metres  a  minute  ? 

Let  ae  be  the  sidewalk,  b  the  position 
of  the  light,  and  cd  one  position  of  the 
man.      Let  ae  =  y,  and  ac  =  x ;  then 
y  —  x :  y : : ■§■ :  4 ;  .'.  dy  =  4j£  dx.    Again, 
let  y  =  ce,  and  x  =  ac  ;  then  y  +  x  :y : : 4  :  | \  .'.dy  =  ^dx. 

5.  The  altitude  of  a  variable  cylinder  is  constantly  equal  to 
the  diameter  of  its  base.  In  general,  its  volume  is  changing  how 
many  times  as  fast  as  its  altitude?  If,  when  its  altitude  is  6 
metres,  it  is  increasing  at  the  rate  of  2  metres  an  hour,  how 
fast  is  its  volume  increasing  at  the  same  instant?  How  fast  is 
the  entire  surface  increasing  at  the  same  instant? 

Ans.  f  ttx2  times,  x  being  its  altitude ;  54  it  kilolitres  an 
hour  ;  36  w  centiares  an  hour. 

6.  The  altitude  of  a  varying  frustum  of  a  right  cone  is  con- 
stantly equal  to  the  radius  of  its  lower  base,  and  the  radius  of 
its  upper  base  is  one-half  that  of  its  lower  base.  If,  when  the 
radius  of  its  lower  base  is  4  metres,  it  is  increasing  at  the  rate 
of  2  metres  an  hour,  how  fast  is  the  volume  of  the  frustum 
increasing  at  the  same  instant? 


APPLICATIONS.  21 

7.  The  area  of  an  equilateral  triangle  increases  hew  many 
times  as  fast  as  each  of  its  sides?  How  fast  is  its  area  increas- 
ing when  each  of  its  sides  is  10  in.,  and  increasing  at  the  rate 
of  3  in.  a  second?  "What  is  the  length  of  each  of  its  sides,  when 
its  area  is  increasing  in  square  inches  30  times  as  fast  as  each 
of  its  sides  is  in  linear  inches? 

Ans.    15  V3  sq.  in.  a  second  •  20V3  in. 

8.  One  end  of  a  ladder  20  ft.  long  was  on  the  ground  5  ft. 
from  the  foundation  of  a  building,  which  stood  on  a  horizontal 
plane,  while  the  other  end  rested  against  the  side  of  the  build- 
ing. The  end  on  the  ground  was  carried  away  from  the  build- 
ing on  a  line  perpendicular  to  it,  at  the  uniform  rate  of  4  ft.  a 
minute ;  how  fast  did  the  other  end  begin  to  descend  along  the 
building?  How  fast  was  it  descending  at  the  end  of  two 
minutes?  How  far  was  the  foot  of  the  ladder  from  the  building, 
when  the  top  was  descending  at  the  rate  of  4  ft.  a  minute? 

Ans.    1.03-1-  ft.  a  minute  ;  3.42  ft.  a  minute  ;  10V2  ft. 

9.  In  the  parabola  whose  parameter  is  8,  the  ordinate 
changes  how  many  times  as  fast  as  the  abscissa?  "What  is  its 
slope  at  any  point  (x,  y)  ?  Find  its  inclination  at  the  points 
whose  common  abscissa  is  ^.  Is  y  an  increasing  or  a  decreas- 
ing function  of  x?    At  what  points  does  the  ordinate  change 

numerically  four  times  as  fast  as  the  abscissa  ? 

4 
In  this  case,  y  is  a  two-valued  function;   and  -  is  -f-  or  — , 

y 

according  as  y  is  +  or  —  ;  .\  the  -j-  value  of  y  is  an  increasing, 
and  the  —  value  a  decreasing,  function  of  x. 

Ans.  -;-;  63°  26' G"  and  116°  33' 54";  Q,  1)  and  (£,  -1). 

10.  In  the  ellipse  cry2  +  L-X2  =  orb2,  the  ordinate  increases 
how  man}*  times  as  fast  as  the  abscissa  ?  y  changes  how 
many  times  as  fast  as  x  at  the  extremities  of  the  axes  of  the 
curve  ?  How  can  the  points  be  found  at  which  y  changes  c  times 
as  fast  as  x?  "What  is  the  slope  of  the  ellipse  at  any  point? 
What,  at  the  extremities  of  its  axes?  Is  y  an  increasing  or  a 
decreasing  function  of  #? 


22  DIFFERENTIATION. 

-£  = —  ;  .'.  when  ?/  changes  c  times  as  fast  as  x, —  =  c. 

dx          ary                                                ,2 ,  al/ 

When  x   and   ?/  have  unlike    signs, is  +>  and  y  is  an 

?*     .              b-x  . 
increasing  function  ;  when  x  and  y  have  like  signs, —  is 

and  y  is  a  decreasing  function. 


cry 


1 1 .  What  is  the  slope  of  f  =  Xs  +  2  a;4  at  (»,  ?/)  ?    What  is  it 

for*=2?  A  3*+8rf  — 

^ns.    ±  -r  ±i«ViO. 

2V»+2x-2  ^ 

12.  What  is  the  slope  of  ?/  =  x3  —  x2  +  1  at  the  point  whose 
abscissa  is  2?    1?   0?    -1?  Ans.    +8;   +1 ;   +0;   +5. 

13.  At  what  point  on  y-  =  2 Xs  is  the  slope  3  ?  At  what  point 
is  the  curve  parallel  to  the  axis  of  x?  Ans.   (2,  4)  ;   (0,  0). 

14.  At  what  angles  does  the  line  Sy—  2x—  8  =  0  cut  the 
parabola  y2  =  8  x  ? 

Find  their  slopes  at  their  points  of  intersection  ;  then  find  the 
angles  between  the  lines  having  these  slopes. 

Ans.  tan-1. 2  and  tan-1. 125. 

15.  One  ship  was  sailing  south  at  the  rate  of  6  miles  an  hour  ; 
another,  east  at  the  rate  of  8  miles  an  hour.  At  4  p.m.  the 
second  crossed  the  track  of  the  first  at  a  point  where  the  first 
was  two  hours  before.  How  was  the  distance  between  the  ships 
changing  at  3  p.m.  ?  How  at  5  p.m.  ?  When  was  the  distance 
between  them  not  changing? 

Let  x  =  the  time  in  hours,  reckoned  from  4  p.m.,  time  after 
4  p.m.  being  -f-,  and  time  before  — .  Then  8a;  and  6  a; -f  12 
will  represent  respectively  the  distances  of  the  two  ships  from 
the  point  of  intersection  of  their  paths,  distances  south  and 
east  being  +,  and  distances  west  and  north  being  — .  Let  y= 
the  distance  between  the  ships ;  then, 

^  =  (8aj)2  +  (6aj+12)2. 

.  (10Q8+72)cto      .  (1) 

9      [64ar  +  (6a;  +  12)2]i  V  ' 


VELOCITY  AND   ACCELERATION.  23 

When  y  does  not  change,  dy=0.  /.from  (1),  100  a; -f- 72  =  0  ; 
.*.#=—  .72  of  60  minutes  =  — 43.2  minutes.  Therefore,  the 
distance  between  them  was  not  changing  at  43.2  minutes  before 
4  p.m.,  or  at  1G.8  minutes  after  3  p.m. 

Ans.    Diminishing  2.8  miles  an  hour;  increasing  8.73. 

33.    Velocity  is  the  rate  of  change  of  the  distance  passed  over 

b}-  a  moving   body.     Hence,   if  s  =  the  distance  and  v  =  the 

ds 
velocity,  v  =  —  (§  30).     If  the  unit  of  s  is  one  foot,  and  the 

unit  of  t  one  second,  v  =  —  ft.  a  second. 
dt 

Acceleration  is  the  rate  of  change  of  velocity. 

Hence,  if  a  =  acceleration,  a  =  —  (§  30). 

dt 


Examples. 

1 .  If  s  =  2  ^,  what  is  the  velocit}'  and  acceleration  ? 

els  o  civ 

Here  v  =  —  =  Qt"  ft.  a  second  ;  a  =  —  =12t  ft.  a  second  : 
dt  dt 

and  the  rate  of  change  of  acceleration  =  —  =  12  ft.  a  second. 

dt 

2.  If  g  =  32.17  ft.,  s  ='^-t2  is  the  law  of  falling  bodies  in 

vacuo  near  the  earth's  surface ;  find  the  velocity  and  accelera- 
tion in  general,  also  at  the  end  of  the  third  and  the  eighth 
second. 

Ans.    a  =  32.17  ft.  a  second  ;  v  =  32.17 1  ft.  a  second  ;  96.51 ; 
257.36. 

3.  Given  s  =  ati,  to  find  v  and  a  in  general,  and  at  the  end 
of  4  seconds. 

Ans.    v  = and  -  ft.  a  second  ;  a  = —   and ft. 

2  V*  4  4V^  32 

a  second  ;  that  is,  the  velocity  decreases  at  the  rate  of =  ft. 

4  V*3 
a  second. 


24  DIFFERENTIATION. 

4.  Given  s3  =  8t2,  to  find  v  and  a  in  general,  and  at  the  end 

of  8  seconds. 

4 
Ans.    v  =  — - — ■   and  -I  ft.  a  second. 
3-Vt 

5.  A  point  moves  along  a  parabola  with  a  velocity  v' ;  re- 
quired the  rates  of  change  of  its  coordinates. 

Since  y2  =  2px,  -^-=*-.  (1) 

dx     y 

If  s  represents  the  length  of  the  curve  traversed,  by  the 
conditions  of  the  problem,  we  have 

—  =  v'.  (2) 

dt  v   ; 

■p.   ,  ds  _dy  m  ds  _ dy  #  Vdar  +  dy2 _ dy    L      dx~      ,0^ 

d£       d£     dy      dt  dy  dt  \        d?/2' 

since  ds  =  Vd*2  +  dy2.  §  16,  Cor.  2. 

Prom  (1),  (2),  and  (3),  we  obtain 


dt\ 


-\-—0,  or-^  = ±- v', 

ir        dt       VjJ2  +  2/2 

which  is  the  rate  of  change  of  y. 
In  like  manner,  we  obtain 

— -  =  — — v',  the  rate  of  change  of  x. 

dt      -y/jr  -f-  y2 

Logarithmic  and  Exponential  Functions. 

34.  The  differential  of  the  logarithm  of  a  variable  is  the  quo- 
tient of  the  differential  of  the  variable  divided  by  the  variable 
itself,  multiplied  by  a  constant. 

Let  y  =  nx;  (1) 

then  dy  —  ndx,  (2) 

and  loga?/*=  loga?i  +  loga#.  (3) 

*  loga<^is  read,  "the  logarithm  of  y  to  the  base  a." 


LOGARITHMIC    AND    EXPONENTIAL    FUNCTION'S.         25 

(4) 
(5) 


From 

(1)  and 
dy     < 

(2), 

X 

i 

From 

(3), 

<2(log, 

.*)' 

=  (Z(log,t#). 

From 

(5)  and 

w, 

1 

d(log( 

«?/) 

c?(log0a?) 

4y 

tte 

y 

a 

Whence  d(\og, 

.«)■ 

bears  the  same 

ratio 

to 

toe   tr\ 

f?-'/         T.nf 

iir  V 

io  tli!  a  rntin  few 

eomp  i 

r»n.T 

—  that  d(log„y) 


y 


as  .r' ;  then  d(\ogax)  =  to  —  when  a  =  #',  and  d(log„?/)  =  m  — 

x  y 

when  y  =  nx' ;  but,  as  n  is  an  arbitrary  constant,  nx'  may  be 

any  number.     Hence,  in  general,  d(\ogay)=m—,  in  which  ra 

is  a  constant.* 

The  constant  ra  is  called  the  Modulus  of  the  system  of  loga- 
rithms whose  base  is  a. 

35.  Let  ra  and  m'  be  the  moduli  of  two  systems  of  loga- 
rithms whose  bases  are  a  and  b  respectively.  If  a  >  b, 
it  is  evident  that  logtt.T  must  change  more  slowly  than  log6x; 
.'.  d(logax)  <  d  (log6rc)  ;  that  is, 

dx    .     .dx  .     , 

m — <  vv  — ,  or  ra  <  ra  . 
x  x 

Hence,  the  greater  the  base  of  a  system  of  logarithms,  the  smaller 
is  its  modulus. 

36.  Naperian  System.  The  system  of  logarithms  whose 
modulus  is  unity  is  called  the  Naperian  system.  The  symbol 
for  the  Naperian  base  is  e. 

*  See  Bice  and  Johnson's  Calculus,  p.  39;  Olney's  Calculus,  p.  25  ;  also 
Bowser's  Calculus,  p.  29. 


26  DIFFERENTIATION. 

The  differential  of  the  Naperian  logarithm  of  a  variable  is, 
therefore,  the  differential  of  the  variable  divided  by  the  variable. 

Thus  we  see  that  Naperiau  logarithms  are  the  simplest  and 
most  natural  for  analytic  purposes ;  and,  hereafter,  the  symbol 
log  will  stand  for  the  Naperian  logarithm. 

37.  The  differential  of  an  exponential  function  with  a  constant 
base  is  equal  to  the  function  itself  into  the  logarithm  of  the  base 
into  the  differential  of  the  exponent,  divided  by  the  modulus  of  the 
system  of  logarithms  used. 

Let  y  =  cx,  then  \ogay  =  x  logac  ; 

dy  (f\ogac 

.-.  m  —  =  logacdx ;    .-.  ffy[=d(c*)]  =  — - — dx. 

Cor.   In  the  Naperian  system,  the  modulus  being  unity,  we 

have 

d((f)  =  <f\ogcdx; 

also,  d(ex)  =  e*dx,  since  loge  =  1. 

38.  The  differential  of  an  exponential  function  with  a  variable 
base  is  the  sum  of  the  results  obtained  by  first  differentiating  as 
though  the  base  were  constant,  and  then  as  though  the  exponent 
were  constant. 

Let  u  =  yx,  then  log  u  =  x  logy ; 

du      .        ,  dy 

.' —  =  log  ydx  +  x-^- ; 

u  *y      ^     y 

.'.  du  =  yx\ogydx  +  xyx~ldy, 
which  is  the  result  obtained  by  following  the  rule  given. 

39.  Logarithmic  Differentiation.  Exponential  functions,  as 
also  those  involving  products  and  quotients,  are  often  more 
easily  differentiated  by  first  passing  to  logarithms.  This  method, 
which  is  illustrated  in  the  two  preceding  demonstrations,  is  called 
logarithmic  d  ifferentiation . 


LOGAlilTHMIC    AND    EXPONENTIAL   FUNCTIONS.         27 


Examples. 


1.   y  =  \og(x*  +  x), 


2.  y  =  loga  Vl  —  Xs. 

3.  /(.r)  =  loga.r3. 

4.  f(x)  =  x\ogx. 

5.  2/  =  clogax. 

G.  /(.r)  =  (loga-)3. 

7.  f(x)  =  af. 

8.  y  =  x*x. 


<ly  =    f  ~*~    clx. 
xr  -\-x 

,  3  mx2      , 

a?/  = ■ clx. 

J      2(^-1) 

...   .      3wi 
/'(*)  =  —• 

/'(a?)  =  logaj+  1. 

c'"g  "*  log  c 

dy  — —dx. 

a  x 

/'(a?)  =  af8(loga;  +  l). 


logy  =  a?  log  a;;  .-.  —  =  af  —  -f-  log x [af  (log a; -f  l)]c/a;. 


.•.  dy  =  x*  x1 

9 .    ?/  =  af\ 

ax-  1 


log*(loga;+l)  +  - 
dy 


dx. 


10.   y 


ax  +  l 


11.    2/  =  log- 


.x   xl  -{-x  lose  a* 

—  _  af  ex  — ' 2 — 

dx  x 


1 


Vl  +  x* 
12.    y  =  log  (logcc). 


13.    y  =  ^ 


dx      x{\  -f-af) 
dx     x  log  a; 
da;-  \x)  V  °Sa; 


11.    ?/ 


Vl  +x 


VT- 


dy 


dx 


O-aOVl-ar" 


In  this  example  and  some  that  follow,  pass  to  logarithms. 


28 

DIFFERENTIATION. 

15. 

xn 
y  ~  (a  +  a;)n' 

dy        naxn~l 
dx      (a  +  x)n+l 

16. 

•7       («2-af')  = 

dy      0     2  or  —  ar  >  ,        ,N, 

-£  =  2se  — ; —  (a-  -f-  ar)  ?. 

da;           (<r— -ar)s 

17. 

y  =  log 

dy  _      1 
da;      1  +  ex 

I 

1 

18.    ?/ =  a-* 


d?/  _  af  (1  —  log  a;) 
da;  ar" 


19.    ?/  =  e*(l-^).  ^  =  e-(i_3.«2_.^). 

da; 


20.   y  =  e 


dy  _ 

4 

da; 

(e'  +  e^)2 

dy  = 
da; 

=  »(f)"(l  +  Iog 

-3- 

d?/  = 

:2a*(ax  +  l)log 

a  da;, 

22.   2/=(aI+l)2. 

23.  Which  increases  the  more  rapidly,  a  number  or  its 
logarithm  ? 

Let  y  =  logua;,  then  dy  =  —dx ;  hence  log0a;  changes  faster  or 
x 

more  slowl}-  than  a;,  according  as  x  <  or  >  m. 

Since,  in  the  Naperian  system,  m  =  1,  dx=  xdy ;  that  is,  the 
number  a;  changes  x  times  as  fast  as  logca\ 

Rem.  The  ratio  of  the  rate  of  change  of  a  number  to  that  of 
its  logarithm  is  variable ;  and  yet  the  hypothesis,  that  it  is  con- 
stant for  comparatively  small  changes  in  the  number,  is 
sufficiently  accurate  for  practical  purposes,  and  is  the  assump- 
tion made  in  using  the  tabular  differences  in  tables  of  logarithms. 

24.  If  y  =  log10a;,  x  changes  how  many  times  as  fast  as  ?/, 
when   x  =  2560,   the  modulus  of  the  Common  system  being 

.434294? 

da;  x          2560    cone    i 
—  =  —  = =  5895  nearly. 

dy     m   .434294         J 


TBIGONOMETBIC    FUNCTIONS.  29 

cl'"C 

25.  By  means  of  the  formula  (£(log.r)  =  — ,  find  d  (x")  in 
which  n  is  any  number,  commensurable  or  incommensurable. 

Let  u  =  xn,  then  log  u  =  n  log  x  ; 

.•.  —  =  n — ,  or  du  =  nxn  'ax. 
u  x 

If  x  were  negative,  to  avoid  logarithms  of  negative  num- 
bers, we  would  square  both  members  of  u  =  xn  before  differen- 
tiating. 

26.  In  like  manner,  obtain  d(xy),  d(xyz),  and  d(x 

27.  Prove  that  d(—  log  ^=^  +  log  c\  =  — — 

\2  a       x  +  a  J      or  — 


x-  —  a~ 


28.    Prove  that  d  [log  (x  +  Va;2  ±  (t2)  +  log  c]  = 


(7a? 


Var'  ±  «2 


29.    What  is  the  slope  of  the  curve  x  =  log10?/,  or  y  =  101? 
What  at  x  =  0  ?     What  at  y  =  5  ? 

^ns.    _J^ — ;  2.3+;  11.51  +  . 
.43429 


Trigonometric  Functions. 

40.  In  the  higher  mathematics,  the  unit  of  angular  measure 
is  the  angle  whose  measuring  arc  is  a  radius  in  length ;  hence, 
if  x  represents  the  length  of  the  measuring  arc  of  any  angle, 

and  r  its  radius,  the  angle  equals  - ;  or,  if  r  =  1,  the  angle  =x. 

r 

In  what  follows  we  shall  assume  r  =  1 . 


41.  Tlie  differential  of  the  sine  of  an  angle  is  equal  to  the 
cosine  of  the  angle  into  the  differential  of  the  angle. 

The  differential  of  the  cosine  of  an  angle  is  equal  to  minus  the 
sine  of  the  angle  into  the  differential  of  the  angle. 


30 


DIFFERENTIATION. 


For,  let  x  represent  any  angle,  or  its  measuring  arc,  and  let 
ac  be  any  value  of  this  arc.  If  at  c  the  mo- 
tion of  the  generatrix  became  uniform  aloug 
the  tangent  cd,  it  is  evident  that  any  simul- 
taneous increments  of  its  distances  from  c 
and  lines  ba  and  bh  may  be  taken  as  the 
differentials  of  the  arc,  the  sine,  and  the 
cosine,  when  x  =  ac  ;  that  is,  if  cd  =  dx, 
ed  =  d(sinse),  and  —  ec  =  d(cosa;).  Now 
angle  edc  =  abc  =  x ;  .'.in  triangle  edc, 
ed[=  d(sin  x)~\  =  cos  xdx, 
and  —  ec[=  d(cos  #)]  =  —  sin  xdx. 

Hence,  as  abc  is  any  value  of  x,  we  have  in  general 
d  sin  x  =  cos  xdx, 
and  d  cos  a?=  —  sin  xdx. 

42.    The  differential  of  the  tangent  of  an  angle  is  equal  to  the 
square  of  the  secant  of  the  angle  into  the  differential  of  the  angle. 
sin  x 


For 


tan  x  ■ 


d  tan  x  = 


coscc 
cos  xd  sin  x  —  sin  xd  cos  x 


COS'iC 

_  (cos2£  +  surx)dx 

cos2x 

dx  2    7 

= =  SQorxdx. 

cos-sc 


43.  Tlie  differential  of  the  cotangent  of  an  angle  is  equal  to 
minus  the  square  of  the  cosecant  of  the  angle  into  the  differential 
of  the  angle. 

For  cot  x 


tan  [ x 

.2 


,dcohx  =  sec2( x\d(  - — x  ]=  —cosec2xdx. 


§42. 


44.  The  differential  of  the  secant  of  an  angle  is  equal  to  the 
secant  of  the  angle  into  the  tangent  of  the  angle  into  the  differential 
of  the  angle. 


For 


'.dsccx  = 


TIlIGONOMETItIC    FUNCTIONS. 

\x 

Bin  a?  da; 


31 


cos  a; 
—  dcosx 


sccxtnnxdx. 


cos-.« 


cos\e 


45.  The  differential  of  the  cosecant  of  an  angle  is  equal  in 
minus  the  cosecant  of  the  angle  into  the  cotangent  of  the  angle  into 
the  differential  of  the  angle. 


For 


cosec  x  ==  sec  f  -  —  x 


\d  cosec  x  =  sec 


—x\ 


:2    J    V2 

=  —  cosec  x  cot  xdx. 


tanf x  )d( x 


§44. 


46.  d  vers  x  =  d(l  —  cos  x)  =  sin  x dx. 

47 .  d  covers  x  =  d(l  —  sin  x)  =  —  cos x dx. 

48.  To  prove  these  theorems  by  the  method  of  limits,  we 
need  the  following  lemma,  which  is  very  useful  also  in  the 
theory  of  curves. 

Lemma.  The  limit  of  the  ratio  of  an  arc  of  any  plane  curve 
to  its  chord  is  unity. 

If  s  represents  the  length  of 
the  curve  mn,  and  pb  =  dx, 
pd  =  ds  ;  and  if  PC  =  Ax',  arc 
pap' = As.  Since  s  is  a  function 
of  x,  we  have 


But 


limit 
Ax=0 

Asl 

AXj 

_ds 
dx 

limit 
A*  =  0 

choi 

•(1  ppr 

_da 

dx 

Fig.  13. 


since 


limit 
Ax-. 


T*Q  [sec  cpp']  =  sec  bpl> 
Hence,  by  division,  we  have 


limit 
AS^O 


As 


chord  pp 


... 


§9. 


32  DIFFERENTIATION. 

Cor.    Since  one-half  of  the  chord  of  an  arc  whose  radius  is 
unit}'  is  the  sine  of  half  the  arc, 


limit  |~sin  x 


=  1. 


49.    To  prove  that  clswx  =  cosxdx  by  the  method  of  limits. 

Let  y  =  sin  x  • 

then  Ay  =  sin  (x  -f  Ax)  —  sin  x. 

But,  from  Trigonometry,  we  have 

sin  a;  —  sin?/  =  2  cos^(x-\-y)  s'm%(x  —  y). 

.'.  Ay  =  2  cos  (x  -J-  £  Ax)  sin  |-  Ax. 

A?/            ,     ,  ,  A   x  sin  &  Ax 
.*.  — u-  =  cos  (x  +  A  Aa;) ^ —  • 

Aa;  V        2       '    ^Aa; 

Now  iin^o  [cos  (^  +  i"  ^03  =  cos  x  > 

»d  2Lmi,o[S-^]=l-  « «,  to. 

.*. -£  =  cos  a;. 

dx 

The  other  theorems  can  be  proved  in  like  manner. 

Examples. 
Differentiate : 

1.  sinaa;.  Ans.  a  cos  axdx. 

0  x  dy  1   .    x 

2.  y  —  cos--  -?-= sin  — 

a  dx  a       a 

3.  y  =  cos a?  =  cos(a^) .  dy=—  3a;2 sin aMc. 

4 .  /(a;)  --  tanm  x  =  (tan  a;)  m.         /'  (a;)  da;  =  m  tan^'a;  sec2 x  dx. 

5.  f(x)  —  tan  x  -+-  sec  a;.  /'  (a;)  =  _lLsin_^. 

cos- a; 

a  •   n        \  dy      cos(lo2a;) 

6.  y  =  sin  (log  a;).  -?-  —  — 4 — - — -• 

dx  x 


TRIGONOMETRIC   FUNCTIONS.  33 

7.  y  =  log(tan»).   •  ^?  =  ^r^ 

8.  y  =  log(sina;). 


dx 

sin  2  a; 

dy  _ 
dx 

cot  X. 

dy  _ 

2 

dx 

sin  2  a; 

9.   y  =  log(cotic). 

10.  y—— — dy  =  —  (cos x -f- sin x) dx. 

sec  a; 

11.  y  =  afeaiax.  dy  =  xn~leBinx  (n  +  x  cos  x)dx. 

12.  y  =  sin(?ia;)sinna;.  dy  =  nsinn~Jx  sin  (nx  +  x)dx. 

13.  y  =  e*  log  sin  x.  dy  =  ex(cota;  +  logsin;c)da\ 

14.  y  =  tan  (log  x) . 

15.  ?/  =  logseca\ 

,  n  cos  x       y  ,      ,      x      dy  1 

1G.    ?/  = £logtan  —      -±= . 

2  sin2  a;  2       dx  sin3  a; 

17.  y  =  4  sinm«a;.  dy  =  4 am  sinm_1aa; cos  aa;efo. 

18.  y  =  x>mx.  -A  —  x*™zi — _ -flog  a;  cos  a:). 

19.  ?/  =  (sinrc)tanx.  ^  =  (sina;)tanx(l+sec2a;logsina;). 

(XX 

20.  y  =  tana*.  jj-  J 

21.  y  =  ^!^-tana;  +  a;.   ^  =  tan4a;. 

22.  y  =  e(°+x),sina;.  ^  =  e(a+I)T2(a+a,-)sina;+cosaf|. 

23 .  y  =  e~aV  cos  ra;.  —  =  —  e~a2x2  (2  a\c  cos  rcc  +  r  sin  rx) . 

24.  2/  =  tanVn^.  gg= -(BeeVT^f. 

*»       2vn^» 


34  DIFFERENTIATION. 

25.  Are  sin  a;,  cosa%  and  tana:  increasing  or  decreasing  func- 
tions of  x? 

d  cos  x  =  —  sin  x  dx ;  and  —  sin  x  is  positive  when  x  is  of  the 
third  or  the  fourth  quadrant,  and  negative  when  x  is  of  the  first 
or  the  second  ;  hence,  cos  x  is  an  increasing  function  when  x  is  of 
the  third  or  the  fourth  quadrant,  and  a  decreasing  function  when 
x  is  of  the  first  or  the  second,  d  tana;  =  sec2xdx,  and  sec2#  is 
always  positive ;    hence  tan  x  is  an   increasing   function   of  x 

between  its  limits  of  continuity;  that  is,  between  x  =  — -  and 

x  —  -,  etc. 
2 

26.  At  what  values  of  x  does  sin  x  change  as  fast  as  a;?  At 
what  values  does  cos  x?  tan  x?  cot  as? 

Ans.    sin  x  does,  when  x  =  0  and  tt  ; 

cos  x  does,  when  x  =  -  and  4tt. 
2 

27.  If  the  change  of  x  and  cos  a  became  uniform  at  30°, 
how  much  would  cosx  decrease  while  x  increases  from  30°  to 
30°  15'? 

Let  y  =  cos  x  ;  then  dy  =  —  sin  xdx  =  —  ^cfa;,  when  a;  =  30°. 

Let  dx  =  15'  =  3-14159=  .004363  ;  then  dy  =  -  .002182. 
180x4 

Hence  cos  .t  would  decrease  .002182.  This  is  evidently  less 
than  the  actual  decrement. 

28.  A  vertical  wheel  whose  circumference  is  20  ft.  makes  5 
revolutions  a  second  about  a  fixed  axis.  How  fast  is  a  point 
in  its  circumference  moving  horizontally,  when  it  is  30°  from 
either  extremity  of  the  horizontal  diameter? 

Ans.    50  ft.  a  second. 

29.  What  is  the  slope  of  the  curve  y  =  sin  a;?  Its  inclination 
lies  between  what  values  ?      What  is  its  inclination  at  x  =  0  ? 

Whatataj  =  -? 
2 

The  slope  =  cos  a; ;  hence,  at  any  point,  it  must  be  something 
between  —1  and  +1  inclusive.     Hence,  the  inclination  of  the 


ANTI-TRIGONOMETRIC    FUNCTIONS.  35 


curve  at  any  point  is  something  between  0  and  -,  or  something 
between  £  it  and  i?  inclusive. 

Ans.    -;  0. 
4 

30.    What  is  the  slope  of  the  curve  y  =  tan  x?     Its  inclination 
lies  between  what  values?     What  is  its  inclination  at  «  =  0? 

What  at  x  =  -? 
4 

Ans.  sec2# ;  between  -  and  -  inclusive  ;  - ;  G3°  2G'  G". 
4  2  4 


Anti-Trigonometric  Functions. 
dx 


50.   c^sin^a)  = 

Vl  —  x2 

Let  y  =  sin"1.^,  then  x  =  sin  y  ; 


.•.  &c  =  cosydy  —  Vl  —  sm2ydy  =  Vl  —  af^cZ?/. 
,\  cZ2/[=d(sin""1a;)]  =  —   ' 

Vl  — a2 

51.  dCcos-1*)  =df|-  sin-1^  = — —  §  50. 

V2  /  VI  —  x2 

52.  d(tan-1x)=    ' 


1  +  z2 
Let  y  =  tan-1  a,  then  x  =  tan  ?/ ; 

.*.  dx  =  sec2ydy  =  (1  -f-  tau2y)dy  =  (1  -f-  ar)dy. 

.-.d?/[=d(tan-1a;)]  =^-^- 

*  To  avoid  the  ambiguity  of  the  double  sign  ±,  we  shall,  in  these  formu- 
las, limit  sin_1x,  cos-'x,  etc.,  to  values  between  0  and  -.     They  mav  be 

made  general,  however,  by  writing  the  double  sign  in  the  second  member 
of  each. 


36  DIFFERENTIATION. 

53.  d(cot_1a;)  =  d( -  —  tan_1a;  )  = — -■ 

K  J        \2  J  1  +  z2 

54.  d(sec1x)  = 


X-yJx1  —  1 
Let  y  =  secr1^,  then  x  =  sec  y ; 


dx  =  sec  y  tan  ydy  =  x-s/x2  —  1  cfy. 
da? 


.*.  efr/[=d(sec_1a;)] 


i»-M  — 


a;  Var'  —  1 

dx 


55.  c7(cosec1a;)  =  d(- —  sec-1aj  j  =  — 

56.  c7(vers_1aj)  = 


xs/a?—  1 
dec 


V2a;  —  x2 
Let  ?/  =  vers" 'a;,  then  x  =  vers  ?/ ; 


.*.  dx  =  sin  y  dy  =  Vl  —  cos2?/  dy 


.\dy[=d (vers  * x) ]  = 


=  Vl  —  (1—  vers?/)2  dy 
=  V2  vers  y  —  vers2?/  <% 
=  V2  a;  —  ar*  c??/. 
dx 


57.   rf(covers_1a;)  =  d(  - — vers  * an  = 


V2  a;  —  ic2 

dx 


J  V2  x  —  x2 

Examples. 


1.    Prove  that  df  sin-1-  )  =  — ■ 

V         «/      Va'-x2 

df-\ 

,(  .  _,a?\  W  dx 

d f  sin  x-    =--;=  =— 


§50. 


ANTI-TRIGONOMETRIO   FUNCTIONS.  37 


2 .    Prove  that  d  ( cos  _1-  )  = 


dfcot~1-i  = -;  dfsec  *-)  = 


adx 


\     '    a)          a'+s8'  V         aJ      x-^Jtf  -  a2 ' 

7/            jO^X                oda;  ,/          ,a;\              dx 

d(cosec~-}  = ;  dfvers    -)  = 

V             aJ         xy/x2  —  a2  V           aJ      V2  ax  -  Xs 


7  /  -1  »\  ^05 

d(  covers    -    =  — 


V  aJ  V2  ax  —  af* 

3.  y  =  a;sin_1a;.  -^  =  sin_1aH ==== 

&c  Vl-a- 

4.  y  =  tana;tan-1a;.  -^=  sec2  a;  tan-1  a;  + 


da;  1+ar2 

x  *     -i    2a;  dy        2(1 -a2) 

5.  w  =  tan  * -•  -£  =  — i L — 

1+ar2  dx      l  +  6ar  +  a;4 

/j  •  -1^4-1  dy  1 

6.  y  =  s\n  * — ! — •  -*£.  =  —  — • 

V2  *"      Vl-'iaj-a8 

_i      1  dy  2 

t.   y  =  sec    - 


2ar— 1  da;  Vl  —  ar 

y  =  cos" ' 


1ar?n—  1  dy  _  _  2  nz""1 


af  n  +  1  dx  ar"  + 1 


9.    y  =  tan-1  (71  tana;).  — "  = 


da;      cos2a;-f-?t2sin2a; 


10.     ys=a*»-\  ^=^n-1x/!iu_^  +  j0^\ 

«*  V     a;  VT^x2/ 

11.  A  wheel  whose  radius  is  r  rolls  along  a  horizontal  line 
with  a  velocity  v' ;  required  the  velocity  of  an}-  point  p  in 
its  circumference ;  also  the  velocity  of  p  horizontally  and 
vertically. 


38 


DIFFERENTIATION. 


Let  apx  (Fig.  14)  represent  the  cycloid  traced  by  the  point  p, 

referred  to  the  axes  ax  and 
ay  ;  then  will  the  horizontal 
and  vertical  velocities  of  p 
be  the  rates  of  change  of  x 
and  y  respectively,  d  be- 
x     ing    the    point    of  contact, 

ad  =  r  vers'1"-     Since  the 


Fig.  14. 


centre  is  vertically  over  d,  its  velocity  is  equal  to  the  rate  of 
increase  of  ad  ; 


d* 
dt 


rvers-1^  ]  = 


r_ dy  ^ 

-\2ry  —  y2  dt 


.'.  -j-  = - — —v'  =  the  velocity  vertically. 


(1) 


Since  ae  =  ad  —  ph,  and  ph  =  Vy{2r  —  y),  the  equation  of 
the  cycloid  is  evidently 

-i# 


x  =  rvers-1-  —  V2  ry  —  y2. 


.'.  dx 


V2  ry  —  y2 


dy. 


dy 


Dividing  bjT  dt,  and  substituting  the  value  of  -^,  we  have 


dx 

dt 


ii         dy     y 

-^  —  -v'  =  the  velocity  horizontally.  (2) 
V  2  ry  —  y2  at       r 


From  §  16,  Cor.  2,  and  equations  (1)  and  (2),  we  have 

ds  =  Veto2  +  dy2  =  (  ry^V  +  jQ  v'dt  =J^v'dt. 
ds        (2v 

''"di=\'v' = the  velocity of  p  •  (3) 

for  the  velocity  of  p  equals  the  rate  of  increase  of  s. 


*  The  symbol  —  indicates  the  operation  of  taking  the  derivative  with 
respect  to  t. 


MISCELLANEOUS    EXAMPLES. 


3U 


From  (1),  (2),  and  (3),  we  have, 
if  2/  =  0,     &=0,     ^'=0,     and^  =  0; 

9         at      '    at  at 


if 


=  r,     -^  =  <y',    — =  y',    and  —  =  V2V; 


dt 


dt 


<lt 


if         w  =  2r,  ^  =  0,     —  =2u',  and  —  =  2v'. 
J  dt  dt  dt 

Hence,  when  a  point  of  the  circumference  is  in  contact  with 
the  line,  its  velocit}'  is  zero ;  when  it  is  in  the  same  horizontal 
plane  as  the  centre,  its  velocity  horizontally  and  vertically  is  the 
same  as  the  velocity  of  the  centre  ;  and  when  it  is  at  the  highest 
point,  its  motion  is  entirely  horizontal,  and  its  velocity  is  twice 
that  of  the  centre. 


Since 


ds 
dt 

ds 


\  r  r 


.'.  —  :  v  : :  V2ry  :  r. 
dt  y 

Hence  the  velocity  of  p  is  to  that  of  c  as  the  chord  dp  is 
to  the  radius  dc  ;  that  is,  p  and  c  are  momentarily  moving 
about  d  with  equal  angular  velocities. 


Miscellaneous  Examples. 
1.    If  y=f(x),  show  that 


limit 


A# 


limit 
A.r  =  0 


/(s  +  A.c)-/(.r)- 


:f(x). 


2.    Find 


If- 


3.    Fmd— H-— ^--1 
dx\_(l  -a2)  J  J 


Ans. 


(a2 -a2)* 

3xr 
(1- 


40  DIFFERENTIATION. 

4.  f(x)  =  (x-S)e2x+4:Xex  +  x  +  3. 

f  (x)  =  (2x -  5)e2l+  4(a;  +  l)e"  4-  1. 

5.  w=iog(vTTa?+vT^a?).  ^-LzlLn^L* 

^         oV  ;  da;        x(l-x*)h 

6.  Find—  [ar5- SlogCl  +  ar5)*]. 


dx1-  '  J  1  +  a3 

„  a;  log  a;  ,  ,     /i        N  dy         losra; 

7.    y  = & — hlog(l—  an.  —  = § — r 

J       1-x         oK  '  dx      (1-x)2 


_  Vx2  +  a2  +  Va,-2  +  62 


Va^  +  a2  —  V^2  +  b2 

dy  _     2x    /q        la;2  -f-  a2         ^ 
<fe~"a?-6\  +  \ar"  +  &2      \a7 


+  P 


+  «'. 
Rationalize  the  denominator  before  differentiating 


n  Var  +  1  4-  a;  dy         fn.2x2-\-\ 

9.   y  =  — •  -2-=2[2x-\ ! — 

Vx2  + 1  —  x  dx        \  Va,-'2  + 1 


10.   y=Ti 


dy  _        —  2a;(2  —  ar) 


(1  +  a2)3  d»      VT=^  V(  1  +  x2)5 


~t\n      dy        ,  i\ :,xn  i  v  1 


11.  y  =  (x-{-VT:-x2)n.     ^  =  n(a;  +  Vl-^)"    ,         „    • 

da;  v  1  —  ar 

12.  2/  =  lo2(Vl+^+VT:^).  ^  =  Ifl J V 

sV  da;      a;V        VH^TV 

_  (sin  war)"1  dy_  ?nw(sin7ia;)m"1cos(ma;  —  nx) 

(cosjm)"  da;  (cos??ia;)"+1 


Vl  +  ar-Vl-ar'  dx          x\       Vl-a;4. 

!  r            .     __i  a;  ,  ,         \x  —  a  dy        2  oar2 

15.    ?/  =  tan  1_  +  log-J_ — •  ^  =  -* i 

a             Mx  +  a  dx     ar  —  a* 


MISCELLANEOUS   EXAMPLES.  41 

,~  _i       a  dy  1 

16.  y  =  sec  l —  •  -*•= 

Va2  —  or2  dx      V"-  —  .'■ 

17.  /(«)  =  (a2  +  a2)  tan  *-•  /'(a?)  =  2 a?  tan"1-  +  a. 

d?/_      a;  sin  _1a? 


18.    y  =-\/l  — oc?  sin*1  x  —  x. 


dx  Vl-if2 


19.    ,/  =  sin  >I=4  ^  = 


1-fa,-2  da;  1  +  ar2 

20.  Find  —  (eaxsmnrx). 

dx 

f  (x)  =  eax  sinm_1  ?•«  (a  sin  ?•#  +  w-r  cos  ?'#) . 

21.  i/  =  log(2a;-l  +  2Vaf'-a--  1).  c^-=  1  ■. 

da;      yV  -  a;  -  1 

22.  y  =  cos^3+'5cosa;.  dy  =       Mx 

5+3cosa;  5+3cosa; 

23.  f{x)  =  e(a+x)2  sin  x.     f'(x)  =  e(a+z)2[2(a  +  a?)  sins  +  cos  a:]. 

24.  f-kg|1og(a +&*>)].  dy^-—-^^——. 

(a  -f-  oa;")  log  (a  -f-  ox") 


25.  y  =  log  ^f-Han-^.  *y  =  T^' 

or  •  -i3  +2  a;  ,  da; 

26.  y  =  sin  ' — ■ dy  =  — — 

Vl3  Vl-Sa.'-ar2 

27.  y  =  exX  tan"aa;.       ^  =  e^T— — ,  +  asxtan-1aj(l  +  logaj) 

dx         [_1  +  ar 


*-y 

28.   a?  =  c~T. 


^^  =  loga;;   .:y=        x        ;  .%&=,       log*      . 
y  1+loga;         dx*      (1 -flog  a;)2 


42  DIFFERENTIATION. 

X2 


29.    y  = 


1  + 


1  + 


1  +  etc.  to  infinity. 
x2  ,  xdx 

y  =  -—-;      .\dy=± 


1  +  y '  Va2  +  i 

i      /  /a 5X  ,        -1%  dy      1    \x  +  a 

a  da;      a;  \  a;  —  a 

Q1  .      Vl  -  x2  +  a;V2        dy      V| 

31       M  =  IOcr— ===: •        —     = ' 

Vl-a,-2  *»      (Vl-ar+a;V^)(l-^) 


32.    y  =  log^— =rl_^ 
^Vl  +  a,-2  —  x 


33 


2/  =  |log^l±^+^  =  log( Vl  +  *?  +  a?)  ; 
Vl  +  ar  —  * 

.  dy  _       1 

'  '  dx     Vl+a,-2 

.      (l  +  af)i  .   ,  .     _i  dy 


VT+x  dx      (l  +  a;)(l  +  ar°) 


CHAPTER   III. 
INTEGRATION. 

58.  A  function  or  variable  is  called  the  Integral  of  its  differ- 
ential. Thus,  a?  is  the  integral  of  Sxrdx ;  and  f(x)  +  C,  C  being 
any  constant,  is  the  integral  of  f'(x)dx. 

Integration  is  the  operation  of  finding  the  integral  of  a 
differential. 

The  problem  of  differentiation  and  the  inverse  problem  of 
integration  ma}-  be  stated  also  as  follows  : 

That  of  Differentiation,  or  of  the  Differential  Calculus,  is,  To 
find  the  ratio  of  the  rates  of  change  of  a  function  and  its  variable. 

That  of  Integration,  or  of  the  Integral  Calculus,  is,  Having 
given  the  ratio  of  the  rates  of  change  of  a  function  and  its  vari- 
able, to  find  the  function. 

The  sign  of  integration  is  J  .     Thus,  J  in  J  4x3cte  indicates 

the  operation  of  integrating  Ax^dx.    Hence,  d  and  J  ,  as  signs  of 

operation,  neutralize  each  other.    For  example,  J  d(a?)  =a,-3,  and 

d  J  3  x"dx  =  3  xrdx.      The  whole  expression  J  ix*dx,  read  "the 
integral  of  4cx\lx,"  represents  the  integral  of  ix\lx. 

59.  Elementary  Principles. 

I.  Since  dC=0,  C  being  any  constant,  J  0=C. 

Hence,  as  0  may  be  added  to  any  differential,  the  general  form 
of  its  integral  will  contain  an  indeterminate  constant  term. 

In  the  Applications  of  the  Calculus,  this  constant  term  is 
eliminated,  or  determined  from  the  data  of  the  problem. 

II.  Since  d{ay  +  ac)  =  ady, 

;.)ady=)d(ay  +  ac)  =  a(y  +  c) 
=  a)d(y+c)  =  ayly. 


44  INTEGRATION. 

Hence,  a  constant  factor  can  be  moved  from  one  side  of  the 
sign  of  integration  to  the  other  without  changing  the  value  of  the 
integral. 

III.    Since  d(x  —  y  +  z  +  c)  =  dx  —  dy  +  dz, 

.*. J  (dx  —  dy  -+-  dz)  =  x  —  y -\-z+c 

=  §dx-jdy  +  §d(z  +  c) 

=  )dx  —  jdy  +  J  dz. 

Hence,  the  integral  of  a  sum  of  terms  is  equal  to  the  sum  of 
the  integrals  of  the  terms. 

60.  Fundamental  Formulas.  Since  integration  is  the  inverse 
of  differentiation,  general  formulas  for  integration  may  be 
obtained  by  reversing  the  general  formulas  for  differentiation. 

1.  f —  =  logic +  logc,*  •.•d(loga;  +  logc)  =  -^- 

2.  (axndx  =  c^  +  C,  ■  rd(c-^  +  C)=axndx. 
J                n+1  \n+l        J 

3.  J  a* logadx  =  ax  +  C,  v  d(a*  +C)  =  a* \ogadx. 
jexdx  =  ex  +  C. 

4.  J  cos xdx  =  since  +  C,  '.'  d(sin a;  +  C)  =  oos&cte. 

5.  J    —sinxdx  =  cosx  +  C. 

6.  J  sec2a;c7a;  =  tana;  +  C7. 

7.  J   —  cosec2xdx  =  cota;  +  (7. 

8.  J  sec  a;  tan  a;  da;  =  seer  -f-  C. 

*  When  the  integral  is  a  logarithm,  it  is  customary  to  write  the  indeter- 
minate constant  term  as  a  logarithm. 


io.  f-  ~  =  S£"x-+  c  =,  £t(>c+n-^)  +e 


a. 


2  2    C   <4x- 


c 


W^-^^c-t^^^c 


*¥■ 


J? 


2J7  /C^i_  =^c,-> 


2C 


2  7  <^c 


i 


77      /"        <?4c 


a 


'^T^&T-  ~~  "  *S<4r'&-  r  c^m4 


*Jtc^ 


*^-ha 


=-  iG^-^/e^-^V 


FUNDAMENTAL   FORMULAS.  45 

9.  J  —  cosec*cot*c?*  =  cosec*  -f-  C 
*10.   I  sin*d*  =  versa;  +  C,  or  —  cos*  +  C. 
til.   J  —  cos * dx  =  covers *  +  C,  or  —  sin*  +  C". 

*13.   f   ~dc    =  cos-1*  +  C,  or  —  sin"1^  +  C". 

H.   (  _«»    =  tan-1*  +  C.  J  aVx1      *  «■ 

*/  1  +  ar 

•15.   f-Zl^£  =  cot"1*  +  C,  or  -  tan"1*  +  C". 
J  1  +  ar 

16.   f     /f       =sec-1*+C7.      J^r^a^i2*0  '~*~K' 

J  x  Var  —  1 

(fa 

■  =  cosec-1*  +  0,  or  —  sec-1*  +  C. 

x  Var'  —  1 


— _— — _ _  =  covers"1*  +  C,  or  —  vers-1*  +  C". 

-\/9  a!  —  ir2 


V2  x  —  x2 


•19 

"  V2 

The  differentials  in  these  nineteen  formulas  arc  the  funda- 
mental integrable  forms,  to  one  of  which  we  endeavor  to  reduce 

*  Two  integrals  having  the  same  or  equal  differentials  must  change  at 
the  same  rate ;  hence  they  must  he  equal,  or  have  a  constant  difference. 
The  constant  difference  between  the  variable  terms  of  the  integrals  in  the 

,   last  four  starred  formulas  is  evidently  - :  for,  when  x  <  -, 

2  2 

cos_1x  +  sin-1x  =  -,    cot_1x  +  tan_1x  =  -,  etc. 
2  2 

The  starred  formulas  are  not  necessary,  since  the  second  integral  in 
each  is  given  by  a  previous  formula. 


46  INTEGRATION. 

every,  differential  that  is  to  be  integrated.  The  processes  of  the 
Integral  Calculus  are  largely  a  succession  of  transformations 
and  devices  to  effect  this  reduction. 

61.  To  facilitate  the  application  of  formulas  1  and  2,  the}' 
may  be  stated  as  follows  : 

I.  The  integral  of  a  fraction  ichose  numerator  is  the  differen- 
tial of  its  denominator  is  the  Naperian  logarithm  of  the  denomi- 
nator, plus  a  constant. 

II.  Whenever  a  differential  can  be  resolved  into  three  factors, 
— viz.,  a  constant  factor,  a  factor  ivhich  is  a  variable  with  any 
constant  exponent  except  —  1,  and  a  factor  which  is  the  differential 
of  the  variable  without  its  exponent, — Us  integral  is  the  product 
of  the  constant  factor  into  the  variable  with  its  exponent  increased 
by  1,  divided  by  the  neiv  exponent,  plus  a  constant. 


Examples. 
Find 

1.  J  aaPdx. 

2.  jbx?dx. 

3.  J2x$dx. 

4.  J*(£a#i-f&Bi)efc. 

7-  f—  2V^  +  C. 

^  s/x 

8.   f(bxi  +  ~)dx.  h—  -  —  +  C. 

J\  a*J  4       Va; 


Ans.  —  +  C. 

7 

f&z§  +  C. 

§tf+a 

axi 

-  bxi  +  a 

-x-t  +  C. 

__6_ 

X2 

+A+a 

EXAMPLES.  47 


f,     Cadx  a ,  ^ 

J    x"  (1  —  n)x"  ' 

10.  CbiGax2  -f  8bxi)^(2ax  +  4bx2)dx. 

Since  a"(6ax,  +  8&»a)  =  (12ax  +  24&x8)efa;,  we  see  that 
the  differential  factor  (2ax-|- 4ox2)<7x  must  be  multiplied 
by  G  to  make  it  the  differential  of  the  variable  Gax2  +  8bxi. 

.'.  f o(6 ax2  +  8 &x*)«(2 ax  +  4 &x*)dx 

=  Hj  (G  ax2  +  8  ox3)  5  ( 1 2  ox  +  24  ox2)  dx 

=  —  (G  ax2  +  8  6a8) !  +  a  §  61 ,  II. 

16 

1 1 .  C[a  (ax  +  ox2)  i  dx  +  2  &  (ax  +  ox2)  5 x  dx] . 
I  [a  (ax  +  bx2)  i  dx  +  2  b  (ax  -f-  ox2)  i  x  dx] 

==  f(aaj  +  &x*)i(a  +  2&»)dx  =  f  (ax  +  ox2)*  +  C. 

12.  f(2a  +  3ox)3dx.  —  (2a  +  3&x)4  +  G. 

J  (a'  +  x8)*  3  ' 

14.  f(l+£x)*dx.  j^.(i +!»)•+ a 

15.  f(6x4-f-2x2-5)(3x2-l)dx.  li^  _  11^ +  5x  + C. 
J  7  3 

1G.   f  c?x   .  log(x-a)  +  logc,  orlog[(x-a)c]. 

J  x  —  a 

17    /y-'dx 
J  a  +  6x" 

/'xn-1dx       1    f%nbxn-1dx       1  ,      ,     .  ,   nN  .  , 
=  —  I  =  — log  (a  +  0xn)  +  logc. 
a  +  bxn      nbJ     a  +  bxH        nb 

C    r)fdx    .  log[(10x3+15)ic]. 


18 

+ 


48 
19 

20 

21 


LNTEGIIATIOX. 


•/; 


2  bx2  dx 


■  f- 

J  8a 


ae  +  bx3 

bxclx 


Gbx2 


log[(ae  +  &a?)?c]. 
log 


2a3  .  6o2  ,   .    ,  ,    o 

ar        ar 


(8a-Gbx?)& 

+  C. 


rt)(2a-x2ydx        5 

22.  f  (&  -  a2)8^^.  fb*xZ  -  f  62a;I  -f  ^fec*  -  ^«¥  +  C 

23.  I  cotada;    =  |  ^— - — -  •  log(csina;) 
J               |_    J      sin  a; 

~  da 

24.  f-*L     =  f-Z. 

J  a;  log;  a;        J  \og;x 


25 

2G 


log  (logs)  +  log c. 
•|ccV2^a;  +  Cr. 


27>   r—(2ax  —  at 

J      (3  aar  —  ar 

f  (2a;4-  3a,-2  +l)*(as8-fa>)efe>.       ^(2a4-  3a,-2  +l)i  +  G, 
I  sin  a;  cos  a; 
J  (logo?) 

31.   fa3*  log  a  da:  [=-£f a^logaSda:]. 

32.J(loga-) 

33.    Ca^dx. 


28 
29 
30 


->,2)cZa; 


:da;. 


da; 
a; 


,da: 
x 


3p  ' 

-£(3  aar* -a-3)  §+(7. 


£  sin2  a;  +  C. 

£  (log  a;)4 -flog  c. 

icP  +  C. 


(logic)m+1  +  logc. 


4  log  a 


a**+C. 


EXAMPLES. 


34. 
35. 
36. 
37. 
38. 
39. 
40. 
41. 
42. 
43. 
44. 
45. 
4G. 
47. 
48. 
49. 


j  endx. 

I  aebxdx. 

j  ccPdx. 

(  cos  (mx)dx\  =—  j  eos(mx)mdx 

j  sec2 \mx)dx. 

j  sin4  a*  cos  a;  dx. 

I  sin3  (2  a)  cos  (2  a;)  da;. 

j  cos4  (3  a;)  sin  (3  a;)  da\ 

f  sec^ar^ardar. 

j  7  sec2  (armada;. 

flog  a;—. 
J  x 

/sin  a;  (fa  _  1  C 

a  +  bcosx  _         6c/ 

j  5  sec(3x)  tan(3a,*)dx. 

I  5  cos  (a  -f  bx)dx. 

j  4  cosec(ax)  cot  (ax)  da. 


—  6  Bin  a;  (fa 

a  +  6  cos  a; 


49 

X 

ne"  +  C. 
sec  0+  C. 


6 


2  lo<x  « 


a21  +  C. 


//< 


sin(mx)  +  C. 


—  tan  (mx)  -+-  C. 
m 

£sinsa;  +  C. 

|sin4(2aO  +  C. 

-T^cos5{3x)+C. 

^tanar3+  C. 

%  tan  xr  +  C. 

£(loga;)2  +  logc. 

log r. 


(a  +  6cos.t)s 
£sec(3a*)  +  C. 


-sin(a-f-  6a;)  -f  C. 


cosec (ax)  +  C. 

a 


50 

50.  j eco"x sinx dx. 

51 .  |  e2einxcosxdx. 

52     HI  +cosx)dx 
J       x  +  sin  a; 

.„   r  xdx  r    i  f 

°  'J  1+4     2J  l+(a^)2_ 
54.   )(1  +x)(l  —x>)x< 


INTEGRATION. 


2  xdx 


—  eC0SI  +  C. 


+  C. 


log  [  (x  +  sin  x)  c] . 


^tarr^  +  C. 


a?      a;4   .   a?  _   x5   ,  „ 
2       T       3        5 


+  1  + 


a?-l 


dx 


5G 
57 
58 


/a;4  da; 
a^  +  1 

/a;"-1  da; 
(ct  +  bxny 

/ba?dx 
3a;4 +  7' 


^-  +  a;+21og(a;-l)+C. 

a;  +  tan-1  a;  +  C. 

3 

6w(l  —  m) 
log[c(3a;4+7)A]. 


62.  Auxiliary  Formulas.  By  integrating  the  equations  in  Ex- 
amples 1  and  2  of  §  57,  and  those  in  Examples  27  and  28  of 
§  39,  we  obtain  the  following  auxiliary  formulas  for  integration  : 

(a)    f-^-2  =  itan-^+C. 
J  x-  +  a1     a  a 


/•u\     C    dx  1  ,      x  —  a,-, 

J  x-  —  a-     2  a       x  +  a 

(°)     j      n r. 

#/  V  a-  —  xr 


+  < 

sin1- -(-  C. 

a 


<»/: 


da; 

s/xT±~a2 


=  log  (a;  -f-  Va*2  ±  a2)  +  logc. 


ATXILIAUY    FORMULAS.  51 

(e)  f — ffe =  lsec  -l-+C. 

J  xy/a?  —  a1     a         a 

(f)  f        dx         =vers1-  +  a 
J  V2  ox*  —  ar*  « 

If  the  differentials  in  these  formulas  were  negative,  we  would 

evidently  have  cos1-,  or  —  sin-1-,  in  place  of  sin-1-; 
a  a  a 

,  ,      x-\- a        ,      a-\- x   .      ,  -,      x  —  a 

and  log — I — ,  or  log — L_,  in  place  of  log ;  etc. 

x  —  a  a  —  x  x-\-  a 

Examples. 

1.  Deduce  formulas   (a),    (c),    (e),  and   (f)  of  §  62,  from 
formulas  14,  12,  1(5,  and  18  of  §  60. 

dx 

(c)     I  =  I  —  =sm  !-  +  C. 

J  Vo7=^     J      L      x*  a 

dx 

,.     f      dx  1  C        a  1       -i^      ~ 

(e)     I —  =-  |  —  =-sec  1-  +  C. 

J  xs/x*  —  a8     «J  »    U2_1      Cl         a 
a  \  as 

2.  Deduce  formulas  (b)  and  (d)  of  §  62  from  formula  1  of 
§  60. 

bmce    — ;= — ; 

x2  —  or     2  a  \x  —  a     x  +  «/ 

/dx  1    f  dx         1    C  dx         1  i     x—a  ,  ^ 

xr  —  or      '2  a  J  x—a      2aJ  x+a      2a       x+a 

To  deduce  (d) ,  assume  Va^iTa2  =  z  —  x. 

.'.  ±  a2  =  z-  —  2xz. 

.  /  x  7  7  dz        dx  dx 

.'.  (z  —  x)dz  =  zdx,   or  —  = 


z       z  —  x      Var2  ±  a2 


52 


INTEGHATION. 


—  =  I  —  =  log  (zc)  =  log  [(x  +  y/x2  ±  a2)c\. 

Var  ±  a2     ^   z 

»/  cr  +  6-ar         &-%/  a2 


7  +  ^2 


ft9 


'  J  Va2  -  6LV         &  ^     la2 

L     \y 

p.    r   dx 

J  a? +16 


—  or 


Ans.  J- tan1—  +  C. 


-sin  x f-  C. 


J  >/< 


da; 


V25  —  a^ 

/•  igdig  r    1  r  2xcix  ~i 

'J  l  +  ic'|_-2J  l  +  (ar)2J 

r       dx         ~_  r 

J  a/(1  —  x—  X2)  J 


-tan-^+C. 
4  4 


sin"1-  +  C. 
5 


dx 


nI-HJ-J 


/: 


dec 


■y/^ax  —  x2) 


f 


dx 


Vf-(-fJ. 


10.  r.* r=  r- 

J  a;2-f  4a;-f  5[_    J  (x+2)2+l_ 

r      <&c       r  _  1  r 

J  X^/h*<r>-  —  ,1-  bJ 


12 


13 


da; 
2a;2  — 2a;  +  l 

da; 


/: 


x^Jb2x2  —  a2 

dx 
V(2a6a;-6V)' 


\  &2 


I  tan"1  a^+C. 

sin  l ! h  C . 

V5 

sin  x 1-  (7. 

a 

tan-^a;  +  2)+0. 
tarr1(2x-l)+C. 


1       _i6a;   ,   n 

-sec  J 1-  C. 

a  a 


-vers  1 h  C. 

&  a 


EXAMPLES. 


53 


J  V8  -4art     3^  V2-(a*)*J  ><  2 

log  /a-  +  -  +  Var2  +  »  +  1  j  +  logc. 


/: 


Vl  +  st?  +  x 


18 


19 


rjMx_ 

/acta 
a'  +  a4 


-fa -fa-2 


20.  r  -fte    . 

■J  V?<  (ca  —  a,-2) 

2i.  r  *»   . 


ilog^rpj+loSc- 

2  a2  a- 

JLtatrl2-^+i  +  0. 
V3  V3 


7l~iCOS_1--f  C. 
C 


vers  ' \-C. 

a 


,_i  2  a; 


Vca*  —  a~ vers-1 f  C. 

2  c 


23. 


24 


Vca;  —  ar 

/'—  xdx  Cc~  ^x  ~  c /7  .  _  f*(c— 2a)(fa    c  f*     da; 

Vca;  —  a~  «^  2  Vca;  —  ar"         «^   2  Vex— af'     2«/  -\/cx— 

f 


dx 


f 


Vaa~  —  & 
V2  aa  -f  ar 


—  logfa-f^|ar 


log  (x+  a  +  V2 «x*  -f-  a--)  -f  logc. 


54 


LNTEG1LAT10N. 


25 


Si 


dx 


x2  +  4x 

r    xdx    . 

J  (a'-x^i 

27     C   Sdx 

J  4-f-9ar 

98    r(&+<,3?)c?a;[ 
J      a2  4-  x2 


ilog 


a;  +  4 


+  logc. 


i-sin-1—  +  C. 
a? 


i-tan-^'-f  C. 
2  2 


29 


30 


31 


/ 


+  ■ 
Srdx 


/: 


Va7^9a;6 
cJa; 


V5  <e4  —  3  a2 

(*      5<fa? 

•^  a;  V3  ar'  —  5 


—  f" 


c&e 


V5 


AT  ""5 


-tan-1-  +  -lose  («2  +  or0)  +  C. 

n.  a.        2 


^vers-^lSa^  +  C. 

V5  sec_1(a;^l- 


iV3 


i(a*|^  +  (7. 


+  C 


82 


c?ar. 


(ar*  —  a2)  5  —  a  sec"1  -  +  6'. 


2    /"(s2  -  a2 
J         x 

r{x2-a2)hdx=  r{a?-dF)dx=  C    xdx  C     a2dx 

J         x  J    x-s/xr—a2      J  -Vxr—ar     J  xVxP—a2 


A  fractional  clifFerential  may  often  be  separated  into  integra- 
ble  parts,  or  reduced  to  an  integrable  form,  by  multiplying  its 
numerator  and  denominator  by  the  same  quantity. 


34 


■y/l-x 


33.  c^i±*dx\  =  cn±m 

J  Vl  -x  J     Vl  -  xr 


s 


sin-1a;-(l-ar)s+(7. 


dx. 


x^/x 


sec  *-  +  log  (x  -f  ^/x2  —  a2)  -+-  log  c. 


35.  r  d*  [=  r  *~3d*  i. 

J  (l-^)'L    J  (ar2-l)ij 


Vl-ar2 


a 


TRIGONOMETRIC    DIFFERENTIALS. 


55 


63.  Trigonometric  Differentials.  The  following  trigonometric 
differentials  are  readily  reduced  to  known  forms.  The  forms 
in  the  first  seven  or  eight  examples  should  be  especially  noted. 


1.  Find 


/; 


dx 
since 


Ans.  log  tan  (-J-a?)  -\-  C. 


/dx   __  r dx rsec2(4-a-)J-r/,i: 
sin  x     J  2  sin  (\x)  cos  (^  x)      J       tan(^-aj) 
=  log  tan  (la)  +  C. 

r  dx  r     r     dx 

J  cos  a;        J    . 


sin  f  -  +  a-* 
V2 


„     r      dx  _  f*sec2xdx 

J  sin  x  cos  x        J      tan  x 


4.  I  cos2xdx\_=  J  (^  +  |  cos  2  a;)  da;]. 

5.  I  sin2  a;  da;. 

e.  fcot^r=fcos*dari. 

J  J     sina; 


logtaug  +  |)  +  C. 


log  tan  x  +  C. 


|  +  £sin2s  +  G. 


--±sin2a;+C. 


tan  a;  da;, 
da; 


sin'.rcos^a; 


log  sin  a;  +  C 

—  log  cos  x  +  0,  or  log  sec  a;  -f  C. 

tan  a;  —  cot  x-\-  C. 


/dx             r(sm2x  +  cos-x)dx      Ct      «    ,  •>  x  -, 

-r-5 —  =  I  - rV o    — =  I  (sec-a;+cosec-a;)da;. 
snr  a;  cos- a;     J         sura;  cos- a;  J 


'xdx. 


I  sin5  s 

j  sin5 x dx  =  j  (1  —  cos2 a;)2 sin x dx 

=  —  I  (1  —  cos2a;)2d(cosa;). 


—  cos  a;  +  -fcos3a;— 


+  C. 


56 


INTEGRATION. 


In  like  manner,  I  sin"  a;  da;  and  i  cosnxdx  can  be  found,  when 
n  is  an  odd  positive  integer. 

■/■ 

■/• 

I  sin5  a;  cos3  a;  da;  =  J  sin5  a;  (1  —  sin2  x)  cos  a;  dx. 

In  like  manner,    I  sinma;cosna;da;  can  be  found,  when  either 
m  or  n  is  an  odd  positive  integer. 

14.  I  sin3 x cos3 x dx.  -^sin4a;  —  isin6a;+  C. 

15.  I  cos4  a;  sin3  xdx.  —\  cos5  x  +  \  cos7  a;  +  C 


10.  I  sin3 a:  da;. 

11.  I  cos3  a;  da;. 

12.  (  cos5a;da;. 

13.  I  sin5 a;  cos3  a;  da;. 


J  cos3  a;  —  cosa;-f  C. 

sin  x  —  -|sin3a;  +  C. 

sin  a;  —  -|  sin3  a; +  ^  sin5  a;  +  C. 

i  sin6  x  —  \  sin8  a;  -+-  C. 


16 


/cos8 
sh 


a;  da; 


( 1  —  sin2  a;)  d  (sin  a;) 
sin4  a; 


1 


,+  C. 


18 


19 


-     /~sin3  x  7 

/.   I  —da;. 

«/  cos- a; 

/sin3  a;  , 
— rda?- 
cosba; 

.   C^dx. 
J  cos' a; 

/siu2aj  , 
— irdx- 
cosba; 

/- — —dx  =  (  tan2 x sec4 a; da;  =  |  tan2  a;  (tan2  a;  +  l)sec2a;da;. 
cos6  a;         J  J  v 


sin  x      3  sin3  a; 
seca;4-cosa;+  C. 

•^sec7a;  —  \  sec5  x  -f-  C. 

isec6a;  —  \  sec4a;  +  C. 

itan5a;  +  ^tan3a;  -+-  C. 


22 


23 


EXAMPLES.  57 

t    ii  sinmaj  7         cosma;   7  ,     .   ,         ,    ,      , 

In  like  manner, ■  dx  or  — dx  ma}' be  integrated,  when 

cos"  a;  sin"  a; 

m  —  n  is  even  and  negative. 

21.  f^^dx.  -W»a>+C. 

J  sin4  a  a 

C^^dx.  UanGx+C. 

J  cos' a; 

— - —  tancc  +  itan3o;  + C. 

cos4  x 

24.  I  tan2a?da;[=  J  (sec2^— l)da;].  tan  a;  —  x  +  C. 

25.  J  cot2#da;.  —  cot#  —  x+C. 

26.  J  tan3#cfcc.  |tan2a;  +  log  cos  a;  -+-  C. 

27.  I  cot^da;.  —  ±-cot2a;  —  log  sin  a;  +  C. 

28.  I  tan5  a;  dx. 

|tan5a;c?a;=  J  (sec2#  — l)tan3#daj=  itan4a;  —  |  tan3#c?a; 
=  ^tan*aj—  j  (sec2 a;—  l)tan<ee?a; 
=  \  tan4  a;  —  |-  tan2  a;  —  log  cos  x-\-  C. 

In  like  manner,  tanmxdx  and  cotma;c?a;  may  be  integrated, 
when  m  is  a  whole  number. 

.   I  tan4  x  dx.  1  tan3  x  —  tan  x  +  x  -f-  C. 

.  J  cot4  acfo.  —  i  cot3  x  -|-  cot  a  -f-  x  +  C. 

.  j  tan6  x  dec.  £tans  a;  —  i  tan3  a;  +  tan  x  —  x  +  C. 


29 
30 
31 


58  INTEGRATION. 

32.      cot6  xdx.  —  ^cot4a;  +  icot2a,*  +  logsina;  +  C. 

64.  Definite  Integrals.  All  the  integrals  jTet  found  contain  the 
indeterminate  constant  term  O,  and  are  called  indefinite  integrals. 

When  C  is  eliminated,  or  determined  for  any  hypothesis,  the 
integral  is  called  a  definite  integral. 

When,  from  the  data  of  a  problem,  we  know  the  value  of  the 
integral  for  some  particular  value  of  its  variable,  C  can  be 
determined.  For  example,  suppose  that  du  =  2axdx,  and  that 
u  =  0  when  x  =  2. 

Since     du  =  2  ax  dx,  u  —  ax2  +  C. 

Since      u  =  0  when  x  =  2,  0  =  4  a  +  C. 

Hence,    C=  —  4  a,  and  u  =  ax2  —  4  a,  a  definite  integral. 

If,  in  an}'  indefinite  integral,  two  different  values  of  the  vari- 
able be  substituted,  and  the  one  result  subtracted  from  the 
other,  C  is  eliminated,  and  the  integral  is  said  to  be  taken  be- 
tween limits.     The  symbol  for  the  definite  integral  of  <jj(x)dx 

between  the  limits  a  and  6  is  I  <j>(x)dx-,  a  and  b  are  called  the 

limits  of  integration,  a  being  the  inferior  and  b  the  superior 

limit.     The  symbol    I     indicates,  that  the  following  differential 

is  to  be  integrated ;  that  a  and  b  are  separately  to  be  substituted 
for  the  variable  in  the  indefinite  integral ;  and  that  the  first  of 
these  results  is  to  be  subtracted  from  the  second. 

In  what  precedes,  we  assume  that  the  integral  is  continuous 
between  the  limits  a  and  b. 

In  the  indefinite  integral,  neither  limit  of  integration  is  fixed 
upon.  In  the  first  form  of  the  definite  integral,  only  the  inferior 
limit  is  determined,  and  the  integral  is  still  a  function  of  the 
variable.  In  the  definite  integral  between  limits,  both  limits  are 
fixed,  and  the  integral  ceases  to  be  a  function  of  the  variable. 

Examples. 

1.  Given  dy  =  (1  +  \ax)*dx ;  find  the  definite  integral  on  the 
hypothesis  that  y  —  0  when  x  =  0. 


EXAMPLES. 


59 


Herey  =  2^(1+faa03  +  <7;  •'•0  =  2f^+C'• 

*      27av       T     '       27a 

2.  Given  dy  =  (x? —  b2x)dx',    find  the   definite   integral,    if 
?/  =  0  when  x  =  2. 

^4ns.  w  = f-  2  &2  —  4. 

J       4         2 

3.  Given  dy  —  — —  ;  find  the  definite  integral,  if  y  =  0 

x      2—x 

when  a?=  1. 

4.  Find    I  nxdx. 


Ans.  y  =  log  (2 a  —  oj'2)  , 


2V  ' 


/nxdx=  -x2  +  C; 
2 

0 


Wc 


=  -  a-  +  C ; 


V+o 


=  ^2+C. 
2 


jTnaxfc  =  |62  +  C  -  go8  +  c)  =  |  (&2  -  a2). 

I  6x?dx. 

|  (aaj2  —  cc3)^. 

Jra    dx 
o  a2  +  X2 

r 


Va2 


24. 
12' 

7T 

4a 

7T 
2' 


9.     I      - —    =  I  (cos0)  "sinOdO 

Jo       cos- 6  \_    Jo 


V2-1. 


*[T+C1 


denotes  the  value  of  -  a:2  +  C  when  x  =  a. 

2 


60  INTEGRATION. 


10 


Jxndx. 
a 

Xco 
e~mdi 


bn+l  —  an+i 


n  +  1 


12-jT      dv- 


V2  r  —  y 

J-ba*  31oa4 

14     r00  xdx  » 

'Jo    1  +  x4'  4' 

siu3j;cos3^f7a;.  — . 

0  12 

Applications  to  Geometry  and  Mechanics. 
65.   Rectification  of  Curves.     From  §  16,  Cor.  2,  we  have 


els  =  -\Zdxr  +  dy2 ; 


This  equation  is  a  general  formula  for  the  rectification  of  airy 
plane  curve ;  that  is,  for  finding  its  length. 

Examples. 
1 .    Rectify  the  semi-cubical  parabola  y2  =  ace3. 

Tj       dy  _  Sax2  #      .  dy2 _  9 one 
dx        2y        '  '  dx2       4 


.-.  s=  C(\  +  ^fdx  =  i  f  (4  +  9  aaj)*tto 


=  (4  +  9^)1     g 

27a  v  ' 

So  long  as  the  point  from  which  s  is  measured  is  undeter- 
mined, C  must  be  indeterminate.  If  the  length  of  the  curve 
be  estimated  from  the  origin,  s  =  0  when  x  =  0. 


AREAS   OF   PLANE   CURVES.  61 

Substituting  these  values  of  s  and  x  in  (1),  we  have 

0  =  -L+C;     .•..,<«+»«>'-»  (2) 

If,  in  (2) ,  a  =  1  and  a  =  f ,  s  =  2^  ;  that  is,  the  arc  of  y=v? 
that  lies  between  the  origin  and  x  =  %,  is  2-y2T  in  length. 

For  the  length  of  the  arc,  the  abscissas  of  whose  extremities 
are  b  and  c,  we  have 

s  =  |  f (4  +  9 ax)Ux  =  (*  +  9«c)»-(4  +  9a6)». 

2.    Find  the  length  of  a  branch  of  the  cycloid 

x  =  r  vers-1— —  V2  ry  —  y2. 
r 

Heref^-X-;    .■Yl  +  ^),  =  VK(Ir-f)-fc 

dy-     2r-y  \       dy-J 

■•■i«(nn*)-«jf(i+g)* 

=  2  V2r  f  (2  r  -  ?/) ~*%  =  8 r. 

Hence  the  length  of  a  branch  is  eight  times  the  radius  of  the 
generating  circle. 

66.   Areas  of  Plane  Curves.     From  §  14,  we  have 

dz  =  ydx ; 

.\z  =  iydx=  \f(x)dx. 

This  equation  is  a  general  formula  for  finding  the  area  included 
between  any  plane  curve  and  the  axis  of  x. 

In  applying  this  formula,  it  must  be  borne  in  mind  that,  area 
above  the  axis  of  x  being  positive,  area  below  it  is  negative. 

For  the  area  between  a  curve  and  the  axis  of  y,  we  evidently 
have 


=jxdy. 


62 


INTEGRATION. 


Examples. 

1 .    Find  the  area  between  y2  =  Ipx  and  the  axis  of  x. 

Here  z  =  J  ydx  =  I  (2px)kdx  =  %x^/2px  +  C=%xy  -f-  C. 

If  the  area  be  reckoned  from  the  origin,  z  —  0 
when  x  =  0  ; 

.-.(7=0,  and  2  =  !-:n/. 

Hence  the  area  oap  =  -|  o apr  ;  and  the  area  of 
the  segment  nop  is  two-thirds  that  of  the  par- 
allelogram MNPK. 

If  oh  =  a,  and  oa  =  b, 

area  bnpd  =  2  I  -\/2pxdx  =  A  V2~p  (5*  —  «f) . 


Fig.  15. 


2.  Find  the  area  of  y  =  ar5  +  &&  between  the  limits  x  =  —  a 
and  x  =  0  ;  also  between  the  limits  a;  =  0  and  x  =  a. 

Ans.  yVa4;  yV<4- 

3.  Find  the  area  of  the  Irvperbola  xy  =  1  between  the  limits 
x  =  1  and  as  =  a. 

Area  =  log  a  ;  that  is,  the  area  is  the  Naperian  logarithm  of 
the  superior  limit.  It  is  because  of  this  property  that  Naperian 
logarithms  are  sometimes  called  hyperbolic  logarithms. 

4.  Find  the  area  inclosed  by  the  axis  of  x  and  the  curve 

y  =  x  —  xz. 

The  inclosed  area  lies  below  the  axis  of  x,  between  x  =  —  1 
and  x  =  0,  and  above  it,  between  x  =  0  and  x  =  1.  These  two 
portions  being  numerically  equal,  the  result  obtained  by  inte- 
grating between  x  =  —  1  and  x=  1  is  0.  To  find  the  required 
area,  obtain  the  area  of  each  portion  separately,  and  take  their 
numerical  sum. 

Ans.  -£. 


AREAS    OF   SURFACES   OF   REVOLUTION.  63 

5.    Find  the  area  of  the  ellipse  ary2  +  b2x2  =  a2b2. 

Area  =  -4  8  Va"  —  x2  dx.     4  |  Vo2  —  x2 dx  =  -a2 ; 
a  Jo  Jo 

for  it  evidently  equals  the  area  of  the  circle  whose  radius  is  a. 


area  =  -7ra2  =  irab. 
a 


6.    Find  the  area  intercepted  between  y2  =  2jpx  and  x2  =  2py. 

Area=  (  V'lpxdx  —  j     — dx  =  -2—- 
Jo        l  Jo    2»  3 


y= 


7.    Required   the   area  intercepted   between  y  = 
x 


and 


1  +  x2 
Ans.  log  4  —  | . 


67.  Since  z=  j  ydx=  (f(x)dx  (§  66),  the  integral  off(x)dx 
can  be  represented  graphically  \>y  the  area  between  the  curve 
y=.f{x)  and  the  axis  of  #.  Hence,  when  j  f(x)dx  cannot  be 
found,  j  f{x)dx  can  be  determined  approximately  by  com- 
puting geometrically  the  area  of  the  figure  formed  by  the  axis 
of  x,  y  =f(x) ,  x  =  a,  and  x  =  b. 


68.   Areas  of  Surfaces  of  Revolution. 

Let  s  represent  the  length  of  the  curve  om, 
and  S  the  surface  generated  by  its  revolution 
about  ox  as  an  axis. 

To  obtain  a  general  formula  for  the  value  of 
S,  let  As  =  arc  pp'  ;    then  AS  =  the  surface  ^ 
traced  by  As. 

limit  f       As 
a*-°  chord  pp' 


Since 


=  i; 


§  48. 


64 


INTEGRATION. 


limit 
As  =  0 


"surface  traced  by 
surface  traced  bj 


j  Asl 
rpprj 


limit 
As  1.0 


AS 
As 


surface  pp' 


As 


,=i; 


limit  f~AS 
As-°[_As_ 


limit 
As  =  0 


2tt  [y  + 


Ay\  chord  pp' 

As 


dS 
ds 


=  2*y. 


S  =  2  7T  Cyds  =  2  7T  (Vl  + C^Y<to. 


When  the  axis  of  involution  is  the  axis  of  y,  we  have,  simi- 
larly, 

S=27rCxds  =  27r  Cxfl  +  — JYefy. 


Examples. 

1 .    Find  the  surface  of  the  sphere. 
Here  the  generating  curve  is  x2  +  y2  =  r2. 


,.S  =  2,jQ 


wi+^Ws. 


dar 


=  2  7r  |  rdx  =  A  ttt2. 


2.    Find  the  surface  of  the  paraboloid  ;  that  is,  of  the  surface 
traced  by  the  revolution  of  a  parabola  about  its  axis. 


VOLUMES    OF    SOLIDS    OF   DEVOLUTION. 


65 


69.   Volumes  of  Solids  of  Revolution. 

Let  V  represent  the  volume  generated  by  the  revolution  of 
op?i  about  ox  as  an  axis.     To  deduce  a  general  formula  for  the 
value  of  V,  let  Ax  =  ab;  then  A?/  =  dp',  and 
A  V  =  the  volume  generated  by  the  revolu- 
tion of  abp'p. 

Now,  volume  abdp  <  AF<  volume  abp'm  ; 

or  -n-y2 Ax<AV<  vr{y  +  Ay)2 Ax  \ 


Ax 

—  =  Try-,  or  V=  -    y-dx. 
ax  J 


Fig.  17. 


(1) 


Or,  to  obtain  (1),  conceive  the  solid  as  generated  by  a  varia- 
ble circle,  whose  centre  moves  along  the  axis  of  the  solid,  and 
whose  radius  is  equal  to  the  ordinate  of  the  generating  curve. 
With  this  conception,  it  is  evident  that,  if  c?£  =  ab,  dV=thc 
cylinder  whose  altitude  is  ab,  and  the  radius  of  whose  base  is  ap. 


.*.  dV=  ~y2dx,  or  V=  tt  j  y2dx. 


When  the  axis  of  revolution  is  the  axis  of  y,  we  have,  simi- 
larly, 

V=tt  j  x2dy. 

Examples. 


1.    Find  the  volume  of  the  prolate  spheroid;  that  is,  of  the 
solid  generated  by  the  ellipse  revolving  about  its  major  axis. 

Here  V=  -  I  ifdx  =  it  I    —  (a2  —  ar)dx  =  |-(2 cnrb2) . 

Hence  the  volume  of  the  prolate  spheroid  is  two-thirds  the 
volume  of  its  circumscribed  cylinder  of  revolution. 


If       a  =  b,  V=^ttci\ 
which  gives  the  volume  of  the  sphere  whose  radius  is  a. 


06  INTEGRATION. 

2.  Find  the  volume  generated  by  the  revolution  of  y3  =  ex 
about  the  axis  of  05,  volume  being  measured  from  the  origin. 

Ans.  V=  f  7r2/2x  =  f-  the  circumscribed  cylinder. 

3.  Find  the  volume  of  the  oblate  spheroid;  that  is,  of  the 
solid  generated  by  the  revolution  of  the  ellipse  about  its  minor 
axis. 

Ans.  V=  f  wtts&  =  |  the  circumscribed  cylinder  of  revolution. 

4.  Find  the  volume  of  the  paraboloid;  that  is,  of  the  solid 
generated  bj*  the  revolution  of  the  parabola  about  its  axis. 

Ans.  V=  \-nxy1,  or  -|  the  circumscribed  cjdinder. 

5.  Find  the  inclosed  volume  of  the  solid  generated  by  the 
revolution  of  y2  —  b2  =  ax*  about  the  axis  of  y. 

70.    From  v  =  —  and  a  =  —   (§  33),  we  have  the  following 
dt  dt 

Fundamental  Formulas  of  Mechanics. 

I.      v  =  —  ;  /.  s  =  j  vdt,  and  t  =  (  — • 
dt  J  J  v 

II.     a  =  —  ;  .'.«=  (  adL  and  t=  {  — • 

dt  J  J    a 

Examples. 

1.  The  acceleration  of  a  moving  body  is  constant;  find  the 
velocity  and  the  distance. 

V=  \  adt  =  at  +  C  =at -{- V0,  (1) 

in  which  v0  represents  the  initial  velocity  ;  that  is,  the  value  of 
v  when  t  =  0. 

.s=  J  vdt=  |  (a.t-\-v0)dt  =  \at-  +  vQt-\-sw  (2) 

in  which  s0  represents  the  initial  distance  ;  that  is,  the  value  of 
8  when  t  =  0. 


APPLICATIONS   TO   MECHANICS. 


G7 


If  the  motion  begins  when  t  =  0,  v0—  0  and  s0  =  0  ;  hence  (1) 
and  (2)  become 

v  =  at      and   s  =  ±at2; 
.  .<  =-*/ —  and  v  =^/2as. 


These  four  formulas  are  the  fundamental  formulas  for  uni- 
formly accelerated  motion. 

The  acceleration  caused  by  gravity  is  32.17+  ft.  a  second, 
and  is  denoted  by  g.  If  we  substitute  g  for  a  in  the  four  for- 
mulas given  above,  we  obtain  the  formulas  for  the  free  fall  of 
bodies  in  vacuo  near  the  earth's  surface. 

2.  By  a  principle  of  Mechanics,  if  ab  be  a  vertical  line,  the 
acceleration  of  a  body  sliding  without  friction  a 
along  the  inclined  plane  AC  is  g  cos  cf>,  in  which 
<j>  =  angle  bac.  Let  s',  v\  and  t'  represent  respec- 
tively the  distance  AC,  the  velocity  acquired  along 
ac,  and  the  time  of  descent;  then,  from  the  for- 
mulas, 

|2s 

we  have 


and  t 


v'=  -\/2gs'  cos</>,  and  t'—\\ 


i  g  cos  <f> 
Let  s  represent  the  vertical  distance  ab  ;  then 


Fig.  18. 


and 


s'  — 


COS0 


tf= 


2s 


g  cos-  <£      cos 


v* =V 2  g  s'  cos  <f>  =  V2^s, 


l      a  s. 


Hence  the  velocity  acquired  by  a  bod)'  sliding  without  fric- 
tion down  ac  equals  that  acquired  by  a  body  falling  verti- 
cally down  ab  ;  and  the  time  of  descent  along  ac  is  —  -  the 
time  of  descent  along  ab. 


cos</> 


08 


INTEGRATION. 


3.    Let  ac  (=  s)  be  the  vertical  diameter  of  airy  vertical  circle 
A  abc  ;  then  the  time  of  descent  from  a  along 

any  chord  ab  ( =  s')  is 


•\  — ■  (Ex.  2),  which  equals  \l — , 

\acos6  \  a 


since  cos  </>  =  —• 
^       s 


Hence  the  time  of  descent  from  a  along  any 
chord  of  the  circle  is  the  same  as  that  along  the  vertical  diameter. 

4.    The  acceleration  varies  directly  as  the  time  from  a  state 

dv 
of  rest  of  the  body;  that  is,  —=a  =  ct;  find  v  and  .s  at  the 

dt 

end  of  time  t. 


Ans.  v  =  \ct2;  s  =  ^cf. 


5.   When  the  velocity  is  a  given  function  of  the  time,  the 
time,  velocity,  distance,  and  acceleration   can  be  represented 
geometrically,  as  follows  : 

Construct  the  locus  of  v  —f(t),  t  being 
represented  by  abscissas,  and  v  by  ordinates, 
the  unit  of  t  being  represented  b}T  the  same 
unit  of  length  as  the  unit  of  v.  Then,  by  §  06, 
the  area  between  the  curve  and  the  axis  of  ab- 


Fig.  20. 


scissas 


equals  i  : 


vdt ;  but,  by  §  7 


0,  s  =  i  vdt. 


Hence,  if  abscissas  represent  time,  and  ordinates  velocities, 
the  area  between  the  curve  v  =f(t)  and  the  axis  of  abscissas 
represents  the  distance  traversed  by  the  moving  body. 

Again,  if  ph  is  a  tangent  at  p,  and  ab  represents  the  unit  of 
time,  dh  represents  the  acceleration  at  the  end  of  the  second 
unit  of  time  ;  for  it  represents  what  would  be  the  increase  of 
the  velochy,  or  ordinate,  in  a  unit  of  time,  if  this  increase  be- 
came uniform  at  the  end  of  the  second  unit  of  time. 

6.  A  body  starts  from  o  (Fig.  21)  ;  its  velocity  in  the  direc- 
tion of  oy  is  constant,  and  in  the  direction  of  ox  is  gt ;  what  is 
its  path? 


APPLICATIONS   TO   MECHANICS. 


69 


Let  ox  and  oy  be  the  axes  of  x  and  y,  respectively ; 
then 


^  =  c,  and  ^  =  gt. 

dt  at    J 


Hence,     y  =  ct,    and  x  =  -J-t2\ 

a      2c2 
9 


0) 


Fig.  21. 


Since  (1)  is  the  equation  of  a  parabola 
referred  to  a  diameter  and  a  tangent  at  its 
vertex,  the  path  of  the  body  is  an  arc  of  a 
parabola.  Hence,  if  it  were  not  for  the  resistance  of  the  atmos- 
phere, the  path  of  a  projectile,  as  a  ball  from  a  rifle,  would  be 
an  arc  of  a  parabola ;  for  its  velocity  would  be  gt  along  the 
action-line  of  gravity,  and  constant  along  the  line  of  projection. 


7.    The  velocity  of  a  body  in  the  direction  of  ox  is  12 1,  and 
in  the  direction  of  oy  is  At2  —  9  ;  find  the  ve- 
locity along  its  path  onm,  the  accelerations  m 
and  distances  in  the  direction  of  each  axis 
and  along  the  line  of  its  path,  and  the  equa- 
tion of  its  path. 

Let  vx,  vy,  v„  and  ax,  ay,  as  represent  respec- 
tively the  velocities  and  accelerations  in 
the  directions  of  the  axes  of  x  and  y  and 
along  the  path,  whose  length  we  will  repre- 
sent by  s. 


Fig.  22. 


Then 


dx 

dt' 


vx—12tj  and 


dy 
dt 


=  4*2-9. 


ds        Ida?  ,  dy'2 


.-.  v3  =  -  =A|—  +  ^-  =  V144 12  +  (4  f  -  9V 
dt      \dt~      dt- 


=  Vlt>*4  +  72£"-f  81  =  4*2-f  9. 


The  accelerations  are 
dv 


dt 


12; 


dv„      0.  ,  dv, 

— 'J-  =  8 1 ;   and  a.  =  — -*  = 
dt  dt 


st. 


70  INTEGRATION. 

The  distances  are 

x=  Cl2tclt=6t2,  (1) 

y=  C(4f-9)dt  =  $P-9t,  (2) 

and  s  =  C(±t2  +  9)dt  =  ±t3  +  9t. 

Eliminating  t  between  (1)  and  (2),  we  obtain   for  the  equa- 
tion of  the  path, 


y 


=  (1—9)  J- 


The  form  of  the  path  is  shown  in  Fig.  22.     (See  Weisbach's 
Mechanics  of  Engineering,  page  148.) 


CHAPTER  IV. 
SUCCESSIVE  DIFFERENTIATION. 

71.  Successive  Derivatives.  Since  /'(#),  the  derivative  of 
f(x),  is  in  general  a  function  of  x,  it  can  be  differentiated.  The 
derivative  of  /'  (x)  is  called  the  second  derivative  of  the  original 
function /(«) ,  and  is  denoted  by /"(#).  The  derivative  of 
f"(x)  is  called  the  third  derivative  of /(#),  and  is  represented 
by  f'"(x)  ;  and  so  on.  /"(#)  represents  the  nth  derivative  of 
f(x),  or  the  derivative  of/"-1  (a). 

Thus,  if 

A")-*!  f>(x)=£(x*)  =  4x>; 

f"(x)  =  —  (4  or3)  =12^:         /'"(a?)  =  —  (12a?)  =  24a; 
^    v  '      dxy      J  v  y      dxy        ' 

/-(a)  =  24;  /v(^)=0. 

f'(x),  f"(x),  f"'(x),  etc.,  are  the  successive  derivatives 
of/(a?). 

72.  Signification  of  f"(x).  Since  /"(#)  is  obtained  from 
/a_1(a;)  in  the  same  way  that/'(#)  is  from  f(x),  the  nth  deriva- 
tive of  a  function  expresses  the  ratio  of  the  rate  of  change  of  its 
(n  —  \)th  derivative  to  that  of  its  variable;  and  the  {n—\)th 
derivative  is  an  increasing  or  a  decreasing  function,  according  as 
the  nth  derivative  is  positive  or  negative. 

Cor.    If  a  is  finite,  and/(a)*=  oo,  /'(a)  =a>,  f"(a)  =oo,  etc. 

For,  when/(a)  =  Go,  f(a-\-h)  is  not  go,  however  small  h  be 

taken.     Hence,  while  x  changes  a  very  small  amount  from  a, 

*/(«)  represents  the  value  of /(x)  for  x  =  a.  The  equation  f(a)=  oo 
means  that/(x)  increases  without  limit,  as  x  approaches  a  as  its  limit. 


72  SUCCESSIVE   DIFFERENTIATION. 

f(x)  changes  an  infinite  amount.  Therefore,  when  x  =  a,  f(x) 
must  change  infinitely  faster  than  x  does;  hence  /'(a)=oo. 
For  like  reason,  if/' (a)  =  oo,  /"(a)  =  oo,  etc. 

Examples. 

1.    Find  the  successive  derivatives  of  ax5-}-  2ar  +  x  +  7. 
Let      /(#)  =  ar3  +  2  ar  +  a;  +  7  ; 

then        f'(x)  =  —  (or3  +  2  Xs  +  x  +  7)  =  3  or  +  4  x  +  1 ; 

/"'(aJ)=£(6aj  +  4)  =  6; 

and       f™(x)=*0. 

2.  Find  the  successive  derivatives  of  ex"'  -\-  ax2  -j-  cZ. 

3 .  If  /(»)  =  a8  log  a?,  prove  that  f™  (x)  =  — 

4.  If  f(x)  —  eax,  prove  that  f"(x)  =  aneax. 

f'l(x)  is  written  out  in  accordance  with  the  law  discovered 
by  inspecting /"(.r)  and  f'"(x). 

5.  If  f(x)=  sinwia,',  prove  that/lv(a)  =  m4 sin mx. 

—  14* 

6.  It  fix)  =  x*  log  «,  prove  that/VI  (x)  =  — 1=. 

ar 

7.  If  f(x)  =af,  prove  that  /"(a;)  =  af  (1  +  log  a;)2  +  af"1. 

8.  If  /(a;)  =  tan  as,  prove  that  /'"  (x)  =  6  sec4 a;  —  4  sec2 a;. 

9 .  If  f(x)  =  log(e*  +  e-*) ,  prove  that  f'"(x)  =  -  8 


(ex+e~xy 


10.    If  /(aj)  =  -^— ,  prove  that  /IV  {x)  = 


l-x   L  "     v  '      (1-x)1 


*  [n,  read  "  factorial  n,"  stands  for  1  X  2  X  3  X4  X  —  X  n. 


SUCCESSIVE   DIFFERENTIALS.  73 

11.    If  f(x)  =  a",  prove  that  /"(a?)  =  (log  a)na*. 

(-i)-y-i 


12.    If  /(a;)  =  log(l  +  x) ,  prove  that  /"(a?)  = 


(l+«)" 


/'(a;)  =  -J—  ;    /"(a?)  =  -I— 0_  ;    finrx\  _  (      *>  l|. 
^  v  y      1  +  a?'   ^    v  '       (1+a?)2'    ^     V  ;      {l  +  x)J 

(-1)3|3  (-ly^ln-l 

f"(x)=- i-%.    .-./»(«)«- 1 \=. 

J    K  }      (1  +  a,-)4  J  KJ  (1  +  a;)" 

13.    If  f(x)  =  ( 1  +  a?)  m,  prove  that 

/»  (a?)  =  «i  (m  —  1)  •  •  •  (m  —  w  +1 )  (1  +  x)  m~n. 

73.  Successive  Differentials.  The  differential  of  the  first  dif- 
ferential of  a  function  is  called  the  second  differential  of  the 
function.  The  differential  of  the  second  differential  is  called 
the  third  differential.  In  like  manner,  we  have  the  fourth,  fifth, 
and  nth  differentials.  d(dy)  is  written  d2y,  and  read  "second 
differential  of  y" ;  d(cPy)  is  written  d?y,  and  read  "third  dif- 
ferential of  y";  and  so  on. 

<^/>  c^Vi  d?y,  etc.,  are  the  successive  differentials  of  y. 

In  differentiating  ?/  =f{x)  successively,  dx  is  usually  regarded 
as  constant ;  that  is,  as  having  the  same  value  for  all  values  of 
x.  This  greatly  simplifies  the  second  and  higher  differentials, 
and  also  the  relations  between  the  successive  differentials  and 
derivatives,  and  is  allowable  ;  for,  when  independent,  x  may  be 
regarded  as  changing  uniformly. 

74.  Relations  between  the  Successive  Differentials  and 
Derivatives. 

dx  dx  \dxj 

If  dx  is  variable,  —  is  a  fraction  with  a  variable  numerator 
dx 

and  denominator,  and  we  have 


74  SUCCESSIVE   DIFFERENTIATION. 

But,  if  dx  is  constant,  -2  is  the  product  of  the  constant  — 
dx  dx 

and  the  variable  dy,  and  we  have 

'"w-sgJ-S-  (2> 

For  the  same  hypothesis  we  have 

/»(.)  -i/ayat  ,>(.)  _g,  /.(.)-  g. 

dx\dx-J      dx*  dx*  dxn 

Hence  dny  =fn(x)dxn ;  whence /"(a;)  is  often  called  the  ni/i 
differential  coefficient  of  /(a,1) . 

For  the  hypothesis  that  dsc  is  constant,  cZ2x=  0,  and  equation 
(1)  becomes  (2),  as  it  evidently  should. 

Examples. 

1.  Find  the  successive  differentials  of  x\ 
Let  y  =  x{ ;  then  dy  =  Ax^dx. 

Differentiating  this  last  equation,  regarding  dx  as  constant, 

we  have 

dry  =  ( 4  da;)  d  (ar)  =  1 2  ar  dar  ; 

.-.  dhj  =  ( 1 2  daf)  d (x2)  =  24 x da? ; 

.:diy  =  24:dxi;  .\d5y  =  0. 

2.  Find  the  successive  differentials  of  5  ar3  +  2  a^  —  3  a;. 

If  y  =  5a;3  +  2 x2  -  3a;,  dfy  =  30da^. 

3.  If  ?/  =  sin  aj,  prove  that  d4y  =  sin  a;  da;*. 

4.  If  y  =  log  (aaj) ,  prove  that  dsy  = -dx*. 

5.  If  y  =  2  a -\/x,  prove  that  —4  = • 

dx8      4  a;* 

6.  If  y  —  log  sin  a;,  prove  that  —*-  = — 

dx3       sin3  a; 


EXAMPLES.  75 

7.  If  y  =  arioso?,  prove  that  —^  =  — 

8.  If  y  =  cosma;,  prove  that  — "  =  m4  cos  ma;. 

a  l  dx> 

9.  If  if  =  2^,  prove  that  ^  =  ^  =  -£^~ 

dar3        ?/5       (2a;)i 

— p— - 
dy  _  P  .      .  drj/  _         dx  _      p2  < 

dec     ?/ '         da?  y1  if ' 

3w-»-  — 
.  dfy  = clx_Spt_    3pi 

"da?  y*  if  ~  (2x)i' 

10.  If  cry2  -f-  £rar  =  a2b2,  prove  that  —  = • 

dar  a2?/3 

11.  If  a2  +  ?/-  =  r2,  prove  that  —4  =  — ■ — 

12.  If  y1  =  sec  2  a;,  prove  that  ?/  H -  =  3  if. 

da? 

13.  If  y  =  e  sin  »,  prove  that  C^(  -  2 ^  +  2 ?/  =  0. 

dar         da; 

14.  Given  s  =  4£3 ;    find  v  the  velocity,  and  a  the  acceleration. 

0.                 ds            dv       d  A?s\      d2s 
Since    v  =  — ,    a  =  — = — [ —    = 

dt  dt       dt\dtj      dt2 

Hence  v  —  —  =  12 12,  and  a  =  —  =  24 1. 

dt  dt2 

15.  If  s  =  ct2  +  bt,  what  is  the  velocit}'  and  acceleration? 

drs 
Ans.  v  =  2  ct  4-  b  ;    a  =  — -  =  2  c. 
cfts 

d2.x      1 

16.  If  if  =  2px,  prove  that  —  =  — 

dy2      p 

Here  a;  is  regarded  as  the  function,  and  dy  is  constant. 


CHAPTER   V. 

SUCCESSIVE  INTEGRATION,  AND  APPLICATIONS. 

75.  The  general  formulas  for  integration  enable  us  to  obtain 
the  original  function  from  which  a  second,  third,  fourth,  or  rath 
differential  has  been  derived. 

For  example,  let  dhj  —  bbdx* ;  then 

d(S)=5bdx'  (i) 


Integrating  (1),  we  have 
dry_ 
dx2 


5  bx  +  d  ;     .-.df^]  =  5  bxdx  +  ddx  : 

\dxj 


.-.cll  =  Ux2+Clx  +  C,; 
dx 

.-.  y  =  %ba?  +  i  dx*  +  C2x  +  C3 


Examples. 

1.  Given  d?y  =  0,  to  find  y. 

Here        f|=0;     r.d(&)  =  0-,     .■.£*=<*; 
dx3  \dx~J  dx2 

.-.df^^C.dx;     ,.cM=C1x+C2; 
\dxj  dx 

.\y  =  i01x2+Osx  +  Ce. 

2.  Given  d*y  =  since  dx*,  to  find  y. 

d(  — ^ )  =  Bmxdx;     .\-pL  =  —  cosaj  +  Ci;  etc. 
\«ar/  cLt3 


PROBLEMS    IX    MECHANICS.  77 


3.    Given  —&  =  3ar  —  ar  ,  to  find  >/. 
dx3 


y  =  Xsc8  +  l-\ogx  -f 1 C^a;2  +  C2x  +  C3. 


r/  ?/ 

4.  Given  — ^  =  sin  a,  to  find  y. 

dx3 

y  =  cosa  +  \  C\x2  +  C2x  +  C8. 

5.  Given  d5y  =  2x~sdx*,  to  find  y. 

y^logx  +  irC^  +  CtX+Cz. 

76.   Problems  in  Mechanics. 

1.  If  the  acceleration  of  a  body  moving  toward  a  centre  of 
force  varies  directly  as  its  distance  from  that  centre,  determine 
the  velocity  and  time. 

Let  fx  =  the  acceleration  at  a  unit's  distance  from  the  centre 
of  force  ; 

x  =  the  varying,  and  c  the  initial,  distance  of  the  body  from 
that  centre  ; 

then  Xfj.  =  the  acceleration  at  the  distance  x. 

tt  ds  dx  ,  *  N 

Here  s=c  —  x:     .'.v  =  —  = :  (1) 

dt  dt  v  ' 

,                                  drs           d-x  /ox 

and  xix  —  a= — -  = -•  (2) 

r  dt2  dt2  v  ' 

Multiplying  (2)  by  —  dx,  we  obtain 

dx  7fdx\  7 

— d  —    =  —  axdx  : 
dt    \dt)         ^ 

...  (£J,w] =_^+a 

Since  v  =  0  when  x  =  c,  C  =  /j.<t  ; 

.'.  v  =  -  —  =  V^Cr-cc2)  ;  (3) 

.♦.  £  =  yu._sCOS-1-.  (4) 

c 


78  SUCCESSIVE  INTEGRATION. 

C  =  0,  since  t  =  0  when  x  =  c. 

If  in  (3)  and  (4)  we  make  x  =  0,  we  have 

v  =  c"v>,  the  velocity  at  the  centre  of  force; 
and  t  =  -|7t/a~5,  |-7r/ir5,  etc. 

Hence  the  time  required  for  the  body  to  reach  the  centre  of 
force  is  independent  of  its  initial  distance  from  that  centre. 

Below  the  surface  of  the  earth,  the  acceleration  due  to  gravity 
varies  as  the  distance  from  its  centre.  Hence,  from  (3)  we  learn 
that  if  a  body  could  pass  freely  through  the  earth,  it  would  fall 
with  an  increasing  velocity  from  the  surface  to  the  centre,  from 
which  it  would  move  on  with  a  decreasing  velocity,  until  it 
reached  the  surface  on  the  opposite  side.  It  would  then  return 
to  its  first  position,  and  thus  move  to  and  fro. 

The  acceleration  due  to  gravity  at  the  surface  of  the  earth 
being  g,  and  the  radius  being  r,  we  have  in  this  case, 

r 

,\  v  =  rV/x  =  V^r,  the  velocity  at  the  centre  ; 

a  o,  -I  l*      q  hip    |20«J  19360 

and  2t  =  Trix  *s=  tt\\—  =  3.141b-\ sec. 

\g  \     32.17 

=  42  min.  13.4  sec, 


which  is  the   time  that  would  be  required  for  a  body  to  fall 
through  the  earth. 

2.  Assuming  that  the  acceleration  of  a  falling  body  above  the 
surface  of  the  earth  varies  inversely  as  the  square  of  its  distance 
from  the  earth's  centre,  find  the  velocity  and  time. 

Let  <r  =  the  varying,  and  c  the  initial,  distance  of  the  body 
from  the  earth's  centre  ; 
r  =  the  radius  of  the  earth ; 

g  =  the  acceleration  due  to  gravity  at  its  surface ; 
a  =  the  acceleration  due  to  gravity  at  the  distance  x. 


PROBLEMS   IN   MECHANICS. 


79 


Then  s  =  c  —  x  ;  and,  from  the  law  of  fall,  we  have 


a  :  g  :  :  r2  :  x2 ; 

_  r/r  _  _  d-x^ 
x2  dt2 


Multiplying  (1)  by  clx,  and  integrating,  we  have 


(1) 


(2) 


If  c  =  oo  ;  that  is,  if  the  body  fall  from  an  infinite  distance 
to  the  earth,  we  have  from  (2) ,  when  x  =  r, 

v  =s/2gr. 
Since  g  =  321  ft.,  and  r  =  3962  miles,  we  have 

v  =  (M&.  x  3962  ]=  6.95+  miles. 
V5280  J 

Hence,  the  maximum  velocity  with  which  a  falling  body  can 
reach  the  earth  is  less  than  seven  miles  per  second. 

From  (2) ,  we  have 

dx 
dt 


"  ,o     ow/c-c-aA*  rtg^hfcx-x2)1* 

-.  dt  =  (_£_)*_=*«*-  =  (^j)le-2x-Gdx ; 
\2  S'ry  Vcaj  -  x2      V2  0T/  2  Vex  -  x? 


c 


ex  — xr)* vers  L  — 

'        2  c 


+  0. 


Since  t=  0  when  x  =  c,  C=  ic7r[ V  ; 

2      \2flrry 


<  = 


W 


ex  — ar)3 vers  x |-ic7r 

2  e 


(3) 


3.  Assuming  that  r,  the  radius  of  the  earth,  is  3962  miles; 
that  the  sun  is  24,000?-  distant  from  the  earth;  and  that  the 
moon  is  60  r  distant ;  find  the  time  that  it  would  take  a  body  to 
fall  from  the  moon  to  the  earth,  and  the  velocity,  at  the  earth's 


80 


SUCCESSIVE  INTEGRATION. 


surface,  of  a  body  falling  from  the  sun.  The  attraction  of  the 
moon  and  sun,  and  the  resistance  of  any  medium,  are  not  to  be 
considered. 

4.  A  body  falls  in  the  air  by  the  force  of  gravity  ;  the  resis- 
tance of  the  air  vai'ying  as  the  square  of  the  velocity,  determine 
the  velocity  on  the  hypothesis  that  the  force  of  gravity  is 
constant. 

Let  /a  =  the  resistance  when  the  velocity  is  unity  ; 

and       *       t  =  the  time  of  falling  through  the  distance  s. 


Then 
and 
Hence 

that  is, 


/dsY 
fx(  —  )  =  the  resistance  of  the  air  for  any  velocity ; 
\dtj 

g  =  the  acceleration  downward  due  to  gravity  alone. 

/  7    \  2 

g  —  fx  [  —  j  =  the  actual  acceleration  downward ; 


d2s 
dt2 


ds 


d 


(ds 

\dt 


d 


(ds 


dt 


ds\2  u  \dt 

—    ,  or  fxdt= i— 

dt)  g 


fi 


(1) 
(2) 


Observing   that   the    second  member  of  (2)   is  of  the  form 
-,  and  integrating,  we  have 


a'  —  x- 


2\gJ 


i    i       i 

#5  +  /*» 


tds 


dt 


i         i  ds 
dt 


(3) 


ds 


(7=0;  since  t  =  0  when  —  \  —  v~\  =  0. 
dt  L       J 

From  (3),  by  principles  of  logarithms,  we  obtain 


.    .      ids 

J  dt 

tds 

J       ^  dt 


—  e2t\^9  • 


ds 
~dt 


problems  in  mechanics. 

•'  —  ij« 


81 


Hence,  as  £  increases,  v  rapidly  approaches  the  constant  value 
r9 


5.  A  bod}*  is  projected  with  a  velocity  v0  into  a  medium  which 
resists  as  the  square  of  the  velocity  ;  determine  the  velocity  and 
distance  after  t  seconds. 

Let  fx.  =  the  resistance  of  the  medium  when  the  velocity 

is  unity ; 

and  s  =  the  distance  passed  over  in  t  seconds. 

Then        /x(  —  )  =  the  resistance  for  any  velocity. 


Hence 


d-s  _    _     fds\2 
dt) 


(U 


1  =  -^ 


ds 
~dt 


—  \x  ds ; 


ds 


.'.  log  —  =  —  fis  +  C  =  —  fj.s  +  log  vQ. 

ds 

C  =  log  v0,  since  —  =  v0  when  s  =  0. 

Hence      —  us  =  log log  v0  =  log  (  —  -*-  v 

dt  V  eft 


■-  v0  =  e  ^s,    or   —  =  — -• 

d*  df      e"s 


(1) 


Hence,  the  velocity  decreases  rapidly,  but  becomes  zero  only 
when  s  =  oc. 

Integrating  (1),  and  solving  the  resulting  equation  for  s,  we 
obtain 

sz=  -log  (/*V0«+1). 
A* 


82 


SUCCESSIVE   INTEGRATION. 


6.    A  body  slides  without  friction  down  a  given  curve;  re- 
quired the  velocity  it  acquires  under  the  influence  of  gravit}'. 

Let  mn  be  the  given  curve,  pa  a 
m  #° 

tangent  at  any  point  p,  and  pa  =  ds  ; 
then  —  pd  =  dy,  and  the  acceleration 
caused  by  gravity  at  p  equals  g  cos  </>, 
in  which  $  =  dpa  (§  70,  Ex.  2). 
drs  ,  dy 

di"  as 


Fig.  23. 


dt     \dt)  J 


■••(|J[-<1— »w+a 


If  .Vo  be  the  ordinate  of  the  starting-point  on  the  curve,  v  =  0 
when  y  =  y0,  and  C=2 gy0. 


v=V2g(y0-y), 


When  y  =  0,  v=v2gy0,  which  is  the  velocity  that  the  body 
would  acquire  in  falling  the  vertical  distance  y0  (§  70,  Ex.  1). 
Hence,  whatever  be  the  curve  down  which,  from  any  point  p,  a 
bod}-  slides  without  friction,  it  has  the  same  velocity  when  it 
reaches  the  line  ox. 


7.    The  base  of  a  cycloid  is  horizontal,   and  its  vertex  is 
downward ;  find  the  time  of  descent  of  a  heavy  bod}'  from  any 

point  on  the  curve  to  its 
vertex. 

Let  the  vertex  o  be  the 
origin  of  coordinates,  y0 
the  ordinate  of  the  start- 
x  ing-point   p,    and   s  the 
length  of  the  curve  reck- 
Then,  from  the  previous  problem,  we 


oned  from  that  point. 
have 


v  =  ^2g(y0  —  y) 


ds 

dt 


dt  = 


ds 


V2a(>o-2/) 


(1) 


PKOBLEMS    IN   MECHANICS.  83 

Since  ds  is  positive,  and  dy  negative, 

The  equation  of  the  cycloid  referred  to  the  axes  ox  and  ov  is 
x  =  r  vera"1-  +  V  2  ry  —  y2. 
^  da  _  f-2  r  —  y\k 

"dy    v    y   / 

.•.*-g+iy*— g)V  (2) 

From  equations  (1)  and  (2),  we  have 


dt  =  — 


2r\h         dy  _       fr\l       dy 


y  J  V2g(y0  —  y)  W  y/y0y  -  f 


^W^  +  c. 


Since  t  =  0  when  y  =  y0, 


(7=f-rvers-12  =  7r/r 


9  J  \9 


...    ^-Yf.-vers-^V 


9  J  \  yo 


t  =  7r  [  —  )  ,  when  y  =  0. 


Hence,  the  time  required  to  reach  the  lowest  point  o  will  be 
the  same,  from  whatever  point  on  om  the  body  starts.  Hence, 
if  a  pendulum  swings  in  the  arc  of  a  cycloid,  the  time  required 


for  one  oscillation  is  2  tta  | —     The  time  of  an  oscillation  being 

independent  of  the  length  of  the  arc,  the  cycloidal  pendulum  is 
isochronal. 


84 


SUCCESSIVE   INTEGRATION. 


Cor.  To  find  the  time  of  descent  along  any  other  curve,  we 
would  obtain  from  its  equation  the  value  of  ds,  substitute  this 
in  equation  (1),  and  integrate  between  the  proper  limits. 

8.  To  find  the  length  and  equation  of  the  Catenary. 
Let  nom  represent  the  form  assumed  by  a  chain,  or  perfectly- 
flexible  cord,  of  uniform  section  and  density,  when  suspended 
from  an}'  two  fixed  points  m  and  n  ; 
then  is  nom  a  catenary.  Let  o,  the 
lowest  point,  be  taken  as  the  origin. 
Let  s  denote  the  length  of  any  arc  ob  ; 
then,  if  <p  be  the  weight  of  a  unit  of 
length  of  the  cord  or  chain,  the  load 
suspended,  or  the  vertical  tension,  at 
b  is  sp.  Let  the  horizontal  tension  be 
ap,  the  weight  of  a  units  of  length  of 
the  chain.  Let  bd  be  a  tangent  at  b  ;  then,  if  bd  represent  the 
tension  of  the  chain  at  b,  be  and  ed  will  represent  respectively 
its  horizontal  and  its  vertical  tension  at  b. 


1 

lv> 

~&/L 

— Ie 

o 

V 

O' 

Fig.  25. 


Hence, 


dy     ed 
dx       BE 

sp      s 
ap      a 

s  I"     dy" 

~\/ds2  —  dx2 
~         dx         ' 

dx           a 

a  [_     dx 

ds       Va2  +  s2 

(1) 


...  x  =  a  C — — —  =  a  log(s  +  Va2  +  s2)  +  C. 


Va2  +  s2 
Since  x  =  0  when  s=  0,  C=  —  aloga. 


alog(-  +  -Jl  +  s 
\a        '        o 


(2) 


Solving  (2)  for  s,  we  have,  for  the  length  of  the  curve  meas- 
ured from  o, 


s  =  -(ea  —  e   °). 
2-  ' 


(3) 


PROBLEMS    IN   MECHANICS.  85 

To  find  its  equation,  we  have,  from  (1)  and  (3), 

7  XX 

dx 

a    ■       -* 
.'.y  =  ^(ea-\-e  a)+C. 

Since  y  =  0  when  x  =  0,  (7  =  —  a. 

is  the  equation  of  the  catenary  referred  to  ox  and  oy. 

If  o'o  =  a,  and  the  curve  be  referred  to  the  axes  o'x'  and 
o'y,  its  equation  will  evidently  be 

x  x 


CHAPTER   VI. 

INDETERMINATE  FORMS. 

77.    When,  for  amT  particular  value  of  its  variable,  a  function 
assumes  any  one  of  the  indeterminate  forms, 


-,     0  •  00,      O0  —  00,     0°,      00°,     l±m, 


the  function,  in  the  usual  sense,  has  no  value  for  this  value  of 
its  variable.  "What  we  call  the  value  of  the  function  for  this 
value  of  its  variable  is  the  limit  which  the  function  approaches, 
as  the  variable  approaches  this  particular  value  as  its  limit. 

Often,  when  a  function  assumes  an  indeterminate  form,  its 
value  may  be  found  by  algebraic  methods. 


1 .    Evaluate 


a?  -  1" 


In  general, 

x'  -  1      ar  +  x  +  1 


Examples. 


;  that  is,  find 


limit 


ag  -  1" 
x2-! 


1 


x+1 


.   limit 
'•  x=l 


'a?  -  1" 


limit 
x=\ 


'x2 + x  +  r 

x  + 1 


2.    Evaluate    (x9~a}\ 
(or  —  a~)  ■» 

In  general, 

(x  —  a  )  a  _  (x  —  a)  A  _  (x  —  a)  A  # 

(or  —  a-) i      (x  —  a) A  (&•  +  a) £       (x  +  a)  ±  ' 


limit  |~  (x  —  q)T 


limit 


ar=a  [_(«  +  <*)* 


=  0. 


EVALUATION   BY   DIFFERENTIATION.  87 

f¥a 


3.    Evaluate   {a2~^+(a-x) 
(a-aj)*  +  (a8-aj8)i_ 


Ans. 


1+«V3 


Evaluation  by  Differentiation. 


78.    To  evaluated,  or  ^*± 


0(a)        0(x)_ 


,  when  it  assumes  the  form 


and 


limit 
Ax  =  0 

limit 
Ax  =  0 

limit 


y(x  +  *x)-f(xy 


A.c 

"<K» 

•  +  Aa-)- 

-<K*) 

Az 

'/(* 

+  A.t)  - 

-/(«)  " 

Ax-°  [_(f>{x+  Arc)  —  0(a) 


=/'(«), 
-*'(»); 

_/'(«>) 
+'(«) 


Ex.  1,  p.  39. 


0) 


Substituting  a  for  #in  (1),  and  remembering  that  /(«)  and 
0(a)  are  each  0,  we  have 


limit 


'/(a  +  Ax)' 

_0(a  +  Aa) 


/'(a)  or  /(a)  =/'(a) 

0'(a)'        <^(a)      0'(a) 


If  /'(«)  =  0,  and  0'(«)  is  not  0,  ^L=  0. 

0(a) 

If  /'(a)  is  not  0,  and  0'(a)  =  0,  ^^  =  oo. 

0(a) 

If  /'(a)  =  0,  and  0'(a)  =  O,  •  ^  '  also  assumes  the  indeter- 
minate form  -  •  Applying  to  it  the  preceding  process  of  reason- 
ing, we  have 

f(a)_f"(a) 


0'(a)      0"(a) 


0 


If  this  also  assumes  the  form  - ,  we  pass  to  the  next  deriva- 

0 
tives,  and  so  on,  until  we  obtain  a  fraction  both  of  whose  terms 

do  not  equal  0. 


88 


INDETERMINATE   FORMS. 


Examples. 


Evaluate 
loaraf 


*— 1 

log£~ 
X  —  1 

1  —  COS  X 


__0 

i    0 


loga:" 
a-1 


=  1. 


78. 


xr 

1  —  cos  x 


x- 


1  —  cos  x 

X1 


sin  x 
2x 


4. 


SB— 1 


e1  —  e~ 
since 

ex  —  e~*  —  2jb 
a;  —  sin  a; 


-4  ns. 


1 


2. 


G. 


a*  —  lf~\ 

X        Jo 

-1 

79.    To  evaluate  f(a) 


-     #— sin 


i      a 
log- 


,  io7tew  it  assumes  the  form  — 
</>(a)  co 


limit  ["/(*) 

_  limit 

1 
1 

_  limit 
x  =  a 

dx 

1    "1 

d 
dx 

1 
_/(*)JJ 

.  limit 


'/(*)" 


aL<K*)J 


limit  fL/Ml2  £( 
*-aU>(*)]2'/'( 


•)J 


;  §78. 


(i) 


EVALUATION   BY   DIFFERENTIATION. 


89 


Since,  when  the  limits  of  two  variables  are  equal,  the  limits 
of  their  equimultiples  are  equal;  we  have  from  (1),  by  multi- 

plying  by  *MZM, 


limit 


7(«0 

.00*0  J 


limit 

x  =  a 


L>'0*)J 


r/(<o=/'(q) 


Cor.    From  §  72,  Cor.,  it  follows  that,  if  a  function  assumes 
the  form  —  for  a  finite  value  of  the  variable,  all  the  functions 

00 

obtained  by  the  formula  given  above  will  also  assume  the  form 
Hence,  to  evaluate  the  function,  it  is  necessary  to  trans- 


00 


form  the  original  function,  or  some  one  of  the  derived  functions, 
so  that  it  will  not  assume  the  form  —  for  this  finite  value  of  the 

00 

variable. 


Examples. 


Evaluate 
.       los:  x 


cosec  x 
I02:  x 


cosec  x 


logic 
cosec  x 

2  sin  x  cos  x 


— cosec  a;  cot  a; 


los:  x 


cot  x 
los:  x 


x  cos  it-    0    cos  x  —  x  sin  x 
Ans.  0. 

00. 


.-!- 


-l. 


logic 

tan  x 
tan  3  it* 


A)is.  0. 


80.   The  forms  0-oo  and  oo  —  oo.   Functions  of  x  that  assume 
the  form  0  • »  or  oo  —  oo  for  a  particular  value  of  a",  may  be  so 


90 


INDETERMINATE   FORMS. 


transformed  that  they  will  assume  the  form  -  or  —  for  the  same 

value   of  x.     Hence  they  can  be  evaluated   by  the   previous 
methods. 


Examples. 


Evaluate 


1.    2* sin  — 


.     a 

sin  — 

a  2X 


Since  2*8^1 — =  — 
2X         2 


sin 


and 


a"l 
=  a  cos  — 

irx~ 
~2~ 

i 

2.    (1  —  x)  tan 


Since   (1  —  x)  tan—  = __, 

2  ,  ttX 

cot  — 


2X  sin 


.), 


and 


1-x 


,    TTX  7T  o  ttX 

cot  —        -  cosec-  — 
2    L    2  2 


(1 —  ic)  tan  — 

v  J         2 


_2 

1       7T 


3.    [sec  x  —  tan  a?]  ^ , 
Since   sec  a;  —  tan  x  -- 


1         sin  x      1  —  sin  # 


and 


1  —  sin  afl 
cos  x    J  i 

_J ; L.1. 

?-l      o.--lJ, 

r— '-— i- 

j_loga;      loga^Ji 


cos  a;      cos  x         cos  a; 
cos  a; 


=  0  ;     .-.  [sec x  —  tan  x~]^  —  0. 

2"- 


Ans. 

2 


EVALUATION   BY   DIFFERENTIATION. 


91 


81.  The  Forms  0n,  oon,  and  1**.  When,  for  any  value  of  »,  a 
function  of  a  assumes  one  of  the  forms,  0°,  oo°,  or  lfc",  its  loga- 
rithm assumes  the  form,  ±0-o>,  and  the  function  is  evaluated 
by  evaluating  its  logarithm  for  this  particular  value  of  x. 


1.    Evaluate  af]0. 


Examples. 


Since    logaf=a:log;a;  = 


and 


log  a; 

2    ~| 

X 

—  (loga;)2 

2  logo; 

X 

—  2af| 

1 

1 

1 

1 

1 

loga>_ 

o               •£ 

0                •%         _ 

0              XT  ^ 

0 

.-.  log  of]0  =  0,  or  af]0  =  1. 

2.    asin*]0. 

bince   logrcsmi=  since  log  x  =  — - — , 

cosec  x 


and 


logo; 
cosec  x 


o     —  cosec  x  cot  cc 
—  2  sin  x  cos  a; 


cc  cos  x 


cos  a;  —  x  sin  cc 
.-.  logasin*]0=0,  orafinx]0=l. 

3.    sina9inx]0.         Ans.  1. 


=  0; 


0; 


4.    sin  astttn  *]£„..       .4ns.  1, 


82.  Componnd  Indeterminate  Forms.  When  the  given  func- 
tion can  be  resolved  into  factors,  some  or  each  of  which  assumes 
the  indeterminate  form,  each  factor  may  be  evaluated  separately. 


Thus,  if  the  function  be  (e*-1)tau'Jj; 

Xs 


,  we  have 


92 


INDETERMINATE   FORMS. 


since 


(ex  — 

1)  tan2  a; 

/'tana 

Xs 

«    \    * 

tan  aT 

=  1 ,  and 

o                             X 

x   J 

=  i; 


=  i. 


Examples. 


tan  x  —  x 
x  —  sin  x 


X  —  Sill  x 


_1 

o    6* 


log(l  +  ^) 


(e*-l)8 


=  2. 


1 

o~3' 


k     1  —  sin  x  +  cos  x 
sin  a;  +  cos  x—1 


n     x2  +  2  cos  a;  —  2" 

iC4 


d    12' 


„  ar8  — 3a;  +  2 

a;1  —  6  ar  +  8  a;  —  3 


log  sin  a; 

9.    xT- 


=  a  log  a. 

5"" 


10.    (cos  ma;)*  J0=  1. 


11.  X'S*  Jo=  oo. 

12.  [astanaj  — ^•7rsecoj]jB.  =  — 1. 


13. 


cos  a;fl  —  cos  ai 


e""°  sin  «( 
2a 


u.    _:+1 


83.  Evaluation  of  Derivatives  of  Implicit  Functions.    If  an 

equation  containing  x  and  y  is  solved  for  y,  y  is  an  explicit  func- 
tion of  a; ;  if  it  is  not  so  solved,  y  is  an  implicit  function  of  x. 


EXAMPLES. 


93 


"When  y  is  an  implicit  function  of  x,  its  derivative,  though 
containing  both  x  and  y,  is  a  function  of  x.  Hence,  when  the 
derivative  assumes  an  indeterminate  form  for  particular  values 
of  x  and  y,  it  can  be  evaluated  by  the  previous  methods. 


Examples. 
1.    Find  the  slope  of  ahf  —  aV  -  x*  =  0  at  (0,  0) . 

tt  dy      2a2x-\-  ix3      0      ,  n 

Here     -r1  =  —  —  =  -,  when  x  =  y  =  0. 


Hence 


dx 

dy 

dx 


2  cry 

_2a2a?  +  4s8' 
o,o  2  cry 


0 


2a2+12ar°" 


dy\2 
dx 


0,0 


1        fty" 
1 ,  or  — 

dx 


°>°        2  a 


=  ±1. 


■Ay 
dx 


dy 

0, 0       ^'^' 


2.    Find  the  slope  of  f  =  ax2  —  Xs  at  (0,  0) . 


Here    * 
dx 


"\dx 


2  aa  —  3  or 
3r 

_  2  rt  —  6  x 
i,o         6y 


2  a  —  6  x 


6y 


dy 
dx 


_  0,0 


2ft  dy 

=  —  =  oo,  or  -& 

o,o     0  dx 


=  ±  oo. 


o,  a 


3.    Find  the  slope  of  a?  —  Saxy  +  f  =  0  at  (0,  0). 


Ans. 


dx 


=  0  and  oo. 


4.    Find  the  slope  of  x*  —  a2xy  +  b2y2  =  0  at  (0,  0) , 


Ans.  & 
dx 


=  0  and 


lr 


5.    Find  the  slope  of  (y2  +  x2)2  —  6 axy2  —  2 ax3  +  a2 a?  =  0  at 


(0,0)  and  («,0). 


Ans. 


dx 


=  ±  oo  ; 


'J  0,0 


dy 


4 


CHAPTER   VII. 
DEVELOPMENT  OF  FUNCTIONS  IN  SERIES. 

84.  A  Series  is  a  succession  of  terms  following  one  another 
according  to  some  determinate  law.  The  sum  of  a  finite  series 
is  the  sum  of  all  its  terms.  An  infinite  series  is  one  the  number 
of  whose  terms  is  unlimited. 

If  the  sum  of  the  first  n  terms  of  an  infinite  series  approaches 
a  definite  limit  as  n  increases  indefinitely,  the  series  is  Con- 
vergent ;  if  not,  it  is  Divergent. 

The  limit  of  the  sum  of  the  first  n  terms  of  an  infinite  con- 
vergent series,  as  n  increases,  is  called  the  Sum  of  the  series. 
An  infinite  divergent  series  has  no  definite  sum. 

85.  To  Develop  a  function  is  to  find  a  series,  the  sum  of 
which  shall  be  equal  to  the  function.  Hence  the  development 
of  a  function  is  either  a  finite  or  an  infinite  convergent  series. 

For  example,  by  division,  we  obtain 
1  —  xn 


1-x 


=  1  +  x  +  ct-2  -f-  ar5  H 1-  xn~x. 


This  finite  series  is  evidently  the  development  of  the  function 

I xn 

for  any  value  of  x. 

Again,  by  division,  we  obtain 

— — =  1  +  x  +  x-  -f  a?  +  x4  H (1) 

When  x  is  less  than  1,  this  infinite  series  is  convergent,  and 
is  easily  proved  to  be  the  development  of  the  function.  But, 
when  x  is  greater  than  1,  the  series  is  divergent,  and  is  not  the 
development  of  the  function;  for,  if  we  put  cc=2,  we  have 
—  1  =  oo,  which  is  absurd. 


taylor's  formula.  95 

86.  Taylor's  Formula  is  a,  formula  for  developing  a  function 
of  the  sum  of  two  variables  in  a  series  of  terms  arranged  accord- 
ing to  the  ascending  powers  of  one  of  the  variables,  with  coeffi- 
cients that  are  functions  of  the  other. 

A  general  symbol  for  an}'  function  of  the  sum  of  x  and  y  is 
f(x  +  y) ,  of  which  (as  +  y)m,  log  (x  +  y) ,  ax+\  sin  (x  +  y) ,  etc. , 
are  particular  forms. 

87.  To  produce  Taylor's  Formula. 

We  are  to  find  the  values  of  A,  B,  C,  etc.,  when 

f(x  +  y)  =  A  +  By  +  Of  +  Dtf  +  Eif  +  •  ••, 

in  which  A,  B,  C,  Z>,  etc.,  are  functions  of  x,  but  independent 
of  y,  the  series  being  finite  or  convergent. 

Let  x'  be  any  value  of  x,  and  A',  B',  C",  etc.,  the  correspond- 
ing values  of  A,  JS,  O,  etc. ;  then  we  have 

f(x'+y)  =  A'+B'y  +  Cy  +  Dy  +  Ey  +  .~;  (1) 

.-.  f(x'+  y)  =  B'+  2  O'y  +  32>Y  +  AEY  +  ».,  (2) 

/"(a;'+2/)=2C'+2.3D'!/  +  34£y+...,  (3) 

f"\x'  +  y)  =  2.3D'+2.S.4:E'y+...,  (4) 

/"r(aj'+y)  =  2.3-4J&'+—,  etc.  (5) 

These  equations,  being  true  for  all  values  of  y,  are  true  when 
y  —  0  (§  6)  ;  hence  we  have 

f(x')  =  A',  f'(x')  =  B\  f'(x')  =  2  Q\ 

/'"(»')  =  [3D',         /"(«')  =  |4  JE?',        etc. 
Solving  these  equations  for  JL',  B',  C,  etc.,  we  have 
^'=/(x'),  B^fW,  c'=^1' 

V-Zp.  *-£g2.  etc. 


96  DEVELOPMENT    OF   FUNCTIONS    IN  SERIES. 

Substituting  these  values  in  (1),  we  obtain 

+rv)^+-  (6) 

Since  the  coefficients  in  (6)  are  equal  to  f(x),  f'(x),  f"(x). 
etc.,  for  x  =  x',  and  since  x'  is  any  value  of  x,  we  have,  in 
general, 

f(x  +  y)  =f(x)  +/'  {x)y  +/»  («)  j£  +/'"  (»)  £ 

+/IV(^)jf+-  (A) 

This  development  of  f(x  +  y)  was  first  published  in  1715  by 
Dr.  Brook  Taylor,  from  whom  it  is  named. 

88.   When  x'—  0,  equation  (6)  of  §  87  becomes 

Ay)  =/(o)  +/(0)y  +/"(0)  |-  +/"'(<>)  £ 

+/IT(0)^-+-  (7) 

in  which  /(0),  /'(0),  etc.,  are  the  values  of  f(y)  and  its  suc- 
cessive derivatives  when  y  =  0. 

Letting  <e  represent  the  variable  in  (7),  we  obtain 
/(*)  =/(0)  +/'(<>)*  +/»(0)  .f  +/'"(0)  jf 

Lf_  L§_ 

+/.v(0)^  +  ...  (B) 

Equation  (B),  called  Maclaurin's  Formula,  is  a  formula  for 
developing  a  function  of  a  single  variable  in  a  series  of  terms 
arranged  according  to  the  ascending  powers  of  that  variable, 
with  constant  coefficients. 


THE    BINOMIAL   THEOREM.  97 

The  completion*  of  Taylor's  and  Maclaurin's  formulas  will  be 
deferred  until  we  have  applied  them  to  the  development  of  a 
few  functions. 

89.  To  develop  (x  +  y)m,  or  to  deduce  the  Binomial  Theorem. 
Here   f(x  +  y)  =  (x  +  y)m ;     .'./(a?)  =  af», 

f'(x)  =  mar-1,  f"(x)  =  m(m  -  l)af-2, 

/'"(»)  =  m(m  -1)  (m  -  2)af-3,  etc. 

Substituting  these  values  in  Taylor's  formula,  we  have 

(x  +  y)m  =  xm  +  mxT-iy  +  m(7^~1)  a-"!--y 

lf_ 
m(ro-l)(m-2)       3  «      _ 

[3_ 

90.  To  develop  logA  (x  +  y ) . 

Here    f(x  +  y)  =  logrt  (x  -f  y)  ;     .-./(a?)  =  logrt  x, 

f'(x)  =  ™,     f"(x)  =  -rn>,     f'"(x)  =  ^,  etc. 
a,*  ar  a; 

Substituting  these  values  in  Taylor's  formula,  we  have 

loga(x  +  y)  =  logax  +  m(y-£+^--f-  +  ..\ 
\x      2  xr      3  ar      4ar  / 

which  is  the  logarithmic  series. 

Cor.    If  #=  1,  and  m  =  l,  we  have 

log(l+»)=f-f  +  f-£+-, 

which  is  the  Naperian  logarithmic  series. 

*  The  series  obtained  by  applying  Taylor's  or  Maclaurin's  formula,  as 
given  above,  to  any  given  function  may  or  may  not  be  the  development  of 
that  function.  Their  complete  forms,  however,  enable  us  to  determine  what 
functions  can  be  developed  by  them. 


08 


DEVELOPMENT   OF   FUNCTIONS    IN   SEEIES. 


91.    To  develop  ax+y. 

Here    f(x)  =  ax,    f'(x)  =  ax  log  a,  etc. ; 


l  +  loga^+(loga)2|l  +  (loga)3|-  +  ... 


92.  To  develop  (a  +  x)m  by  Maclaurin's  Formula. 
Here      f(x)=  (a+x)m ;  .-.    f{0)  =  am. 

f(x)  =  m(a+x)m-1;  .'.  /'(0)  =  mam~x. 

f"(x)  =  m(m  —  l)(a+x)m-2;  .-.  /"(0)  =  m(m  —  l)am  2. 
etc.  etc. 

Substituting  these  values  in  Maclaurin's  formula,  we  have 
(a  +  x)m  =  am  +mam-1x  +  m(^  —  1)  cT^ar3  -) 

93.  To  develop  sin  x. 
Here       /(#)  =  sin  a  ; 

/'  (jc)  =  cos  x  ; 
f"(x)  =  —  sin#; 
/"'(#)  =  —  cos  a;; 
f"(x)  =  sin  x  ; 
fv(x)  =  cos  a;; 
etc. 


;.   /(0)  =  o. 
••  /'(0)  =  i. 

••  /"(0)  =  0. 

••/'"(0)  =  -i. 

••/IV(0)  =  o- 
•.  r(0)  =  i. 

etc. 


Substituting  these  values  in  the  formula,  we  have 

/v***  /y&  /yd  /v1" 

•  .>  ,i  ,(  ,< 

sincc  =  #  —  ■ K , ••• 

li      li      H     li 

94.    To  develop  cos  x. 


cos  x=l 


2_      l£_      |6_      [8_ 


This  result  could  be  obtained  by  differentiating  the  value  of 
sin#  found  in  §  93. 


THE  LOGARITHMIC    SERIES.  99 

95.  To  develop  ax. 

a'  =  1  +  log  a  j  +  (log  a)8—  +  (log  a)3—  + ..., 

•which  is  the  exponential  series. 

Cor.  1.    If  a  =  e,  the  Naperian  base,  we  have 

a-sssi+2+i£+.££ +  «£+... 
1    [L    IL    IL 

Cor.  2.    Putting  a;=l,  we  have 

|_2_      1 3       |4       |5 
Hence  e  =  2.718281  +  . 

96.  To  develop  log&  (1  +  x) . 

iog.(i+»)=«(«-Y+y-Y+y--"-)         0) 

which  is  the  common  logarithmic  series. 
If  m  =  1 ,  we  have 

log(l  +  a!)  =  aJ-^  +  ^-^  +  |— ...,  (I*) 

which  is  the  Naperian  logarithmic  series. 

In  §  104,  this  development  is  proved  to  hold  only  for  values 
of  x  between  —1  and  +1  ;  hence,  in  this  form,  it  is  useless  for 
the  computation  of  Naperian  logarithms  of  numbers  greater 
than  2.  We  therefore  proceed  to  adapt  the  Naperian  logarith- 
mic series  to  the  computation  of  logarithms. 

Substituting  —  x  for  x  in  (1'),  we  have 

log(l-aj)  =  -»-  — ------ (2) 

oV  ;  2        3        4        5  V    ' 

Subtracting  (2)  from  (l'),  we  have 

log(l  +  x)-log(l-^)  =  2^  +  ^  +  ^-f-^  +  ...^(3) 


100  DEVELOPMENT    OF   FUNCTIONS   IN   SEEIES. 

Let  x  = ;    then  =    "*"    ;    and,  for  any  positive 

ZZ  •+  Y  I  —  X  Z 

value  of  z,  #<  1. 

Hence      log(l  +  a)  —  log(l  —  a-)  =  log(z +  1)  —  logz. 
Substituting  these  values  in  (3),  we  have 
log(z  +  l)  —  logz 

2z+l+3(2z+l)3+5(2z+l)5+'"/ 

Equation  (4)  is  true  for  any  positive  value  of  z  ;  and,  since 
the  series  converges  rapidly,  log(z+l)  can  be  readily  computed 
when  logz  is  known. 

Putting  z  =  1  iu  (4) ,  we  obtain 

log2  =  2fI  +  -1      ■      1      •      > 


,3      3-33      5-35      7-37 
Summing  six  terms  of  this  series,  we  find 

log  2  =  0.693147  +  . 
Putting  z  =  2  in  (4) ,  we  have 

log3=log2  +  2(l+—  +  — +  _?_-}-. ..^ 
&  \5      3-58     5-55     7-57         J 

=  1.098612  +  . 
log  4  =  2  log  2  =  1.386294+ . 
Putting  z  =  4,  we  obtain 

log  5  =  log  4  +  2 1  -  + 


/I  +  J_+J_  +  J_+...>| 

\9      3-93      5-95     7-97        / 


=  1.6094379+. 
log  10  =  log  5  +  log  2  =  2.302585 +  . 

In  this  way  we  can  compute  the  Naperian  logarithms  of  all 
numbers. 

Cor.  1.    Letting  m  and  m'  be  the  moduli  of  two  S3'stems  of 
logarithms  whose  bases  are  a  and  a',  from  (1)  we  evidently 

have  log„(l  +  3Q  =  m^  ,5v 

logu.(l  +  a)      m'' 

in  which  x  lies  between  —1  and  +1. 


THE   LOGARITHMIC    SERIES.  101 

To  prove  that  the  principle  in  (5)  is  true  for  all  numbers,  let 
loga  ( 1  +  a;)  =  «,  and  log„.  ( 1  +  x)  =  w  ; 

then  au  =  (1+  a?)  =  a'w,  or  a'=  a" ;  (6) 

and  log,,(l  +  *)=ii.  (7) 

log„.(l  +  aj)       ^y 

Again,  y  being  any  number,  let 

\ogay  =  z,  and  log,,*//  =  v  ; 

then  az  =  y  =  a",  or  a'=  a7 ;  (8) 

(9) 


and              **-*«« 

logu>y     ^ 

From  (6)  and  (8), 

«                 z 

a*=scF,  or  — 

_z 

?o 

w 

(11) 


(10) 

From  (7),  (9),  (10),  and  (5),  we  have 

log,?/  __  log,  (1+aQ  =  m < 
loga.?y     log„»(l  +  aj)      m' 

Hence,  the  logarithms  of  the  same  number  in  different  systems 
are  proportional  to  the  moduli  of  those  systems. 

Con.  2.    If,  in  (11)  of  Cor.  1,  we  let  a'=  e,  we  have 
m'=l, 
and  loga?y=  mlogy.  (12) 

Hence,  the  logarithm  of  a  number  in  any  system  is  equal  to 
the  Naperian  logarithm  of  the  same  number  multiplied  by  the 
modulus  of  that  system. 

Cor.  3.    If,  in  (12)  of  Cor.  2,  y  =  a,  we  have 

1 

m  = 

loga 

Hence,  the  modulus  of  any  system  of  logarithms  is  the  recipro- 
cal of  the  Naperian  logarithm  of  its  base. 


102  DEVELOPMENT   OF   FUNCTIONS   IN   SERIES. 

In  the  Common  system,  a  =  10  ; 

hence  m  = = =.434294  +  . 

log  10      2.302585 


97.  Taylor's  formula  evidently  fails  to  develop  f(x  +  y)  fol- 
ic =  a,  if  /(a)  or /"(a)  is  infinite,  while  n  remains  finite;  while 
Maclaurin's  fails  to  develop  f(x)  for  any  value  of  x,  if  /(0)  or 
/"(0)  is  infinite  for  n  finite. 

For  example,  by  Taylor's  formula,  we  have 

(aj-6+«)i=(aj-6)H V- — y- p..  (1) 

When  x  —  6,  (1)  becomes 

Vy  =  00  —  CO  -( 

Hence  the  formula  fails  to  develop  (x  —  b  +  y) '  for  a;  =  6. 
By  Maclaurin's  formula,  we  have 

logX=  —   GO  +  00   —   CO   +  ... 

Hence  Maclaurin's  formula  fails  to  develop  log  x  for  any  value 
of  x. 

98.  To  complete  Taylor's  and  Maclaurin's  formulas  so  that 
they  shall  enable  us  to  determine  what  functions  can  be  devel- 
oped by  them,  we  need  the  following  lemma  : 

Lemma.  If  f(x)  is  continuous  between  x  =  a  and  x  =  b,  and 
iff  (a)  =  f  (b)  =  0,  then  f'(x),  if  continuous,  must  equal  zero  for 
some  value  o/x  between  a  and  b. 

For,  if  f{x)  is  continuous,  and  f(a)  =f(b)  =  0  ;  then,  as  x 
changes  from  a  to  b,  f(x)  must  first  increase,  and  then  decrease  ; 
or  first  decrease,  and  then  increase;  hence  f'(x)  must  change 
from  +  to  — ,  or  from  —  to  +,  and  therefore,  if  continuous, 
pass  through  0. 


taylok's  and  maclaurin's  formulas.        103 

99.  Completion  of  Taylor's  and  Maclaurin's  Formulas.  To  be 
the  development  of  f(x+y),  the  series  in  Taylor's  formula  must 
be  finite  or  infinite  and  convergent  (§  80). 

vn 
Let  P"f—  be  the  difference  between  f(x  +  y)  and  the  sum  of 
[n 

the  first  n  terms  of  the  series ;  then  we  have 

/(*  +  y)  =/(*)  +/»(■)  f  +/'(*)  £  +/'»(*)  i 

1  \2_  \3_ 

+  ...+/^«)-Cl  +  plC.       (l) 

n  —  1  n 


We  proceed  to  find  the  value  of  P. 

Letting  y  =  X  —  xin  (1),  and  transposing,  we  have 


f(X)  -/(*)  -/'(*)  *p  -/»(*)  X^Z*1 


-/"'(*)  ^=^-3 r~\x)  ^4^r 

|o  |n  —  1 


_F(x-xr= 


0.  (2) 


Let  .F(z)  represent  the  function  of  z  obtained  by  substituting 
z  for  x  in  the  first  member  of  (2)  ;  then 

F(z)  =/{X)  -f(z)  -f'(z)  *=*  -f"(z)  {X~Z)* 

p-\*){X~Z\n~l  -  P(X-z)n-       (3) 

\n  —  1  \n 

Substituting  X  for  z  in  (3),  we  obtain  F(X)  =  0. 

From  (2)  we  see  that  the  right-hand  member  of  (3)  is  0  for 
z  =  x  ;  hence,  by  substituting  x  for  z  in  (3),  we  obtain  F(x)  =  0. 

Differentiating  (3)  to  obtain  F'(z),  we  find  that  the  terms  of 
the  second  member  destroy  each  other  in  pairs,  with  the  excep- 
tion of  the  last  two,  and  obtain 


10-4  DEVELOPMENT    OE   EUNCTIONS   IN   SERIES. 

\n  —  1  \n  —  1 

Whence,  P=fn(z)  when  2?"(z)  =  0.  Since  .F(z)  =  0  when 
z  =  X,  and  also  when  z  =  x,  F'(z)  =  0  for  some  value  of  z 
between  X  and  a  (§  98).  Now,  bj-  giving  to  0  some  value  be- 
tween 0  and  +1,  an}-  value  between  x  and  X  can  evidently  be 
represented  b}-  x  -f-  0(X  —  x)  . 

Hence,     P=fn[x  +  6{X -  x)~\=fn{x  +  Oy) . 
Substituting  this  value  of  P  in  (1),  we  have 

|n  —  1  |n 

which  is  one  complete  form  of  Taylor's  formula. 

Cor.    Letting  x  =  0,  and  putting  x  for  y,  we  have 

f{x)  =/(0)+/'(0)s  +/"(0)  -|  +/"'(0)  ^  +  .- 

|?&  —  1  |n 

which  is  one  complete  form  of  Maclaurin's  formula. 

100.  If  we  had  let  Pxy  be  the  difference  between  f(x-\-y)  and 
the  sum  of  the  first  n  terms  of  the  series  in  Taylor's  formula, 
we  should  have  found 

Pl=r(x+ey)yn-1^-o)n-\ 

\n  —  1 


Hence  a  second  complete  form  of  Taylor's  formula  is 


n  —  1  71—1 


TAYLOifs  a>;d  maolauiiin's  FORMULAS.  105 

Cou.  1.  The  corresponding  complete  form  of  Maclaurin's 
formula  is  evidently 

/(*)=/(<>)  +/'(0),;  +/"(0).-y-  +•■■+/"  Wr^T 

+/"(^.t)(176)"i'j:"' 

I  n  —  1 

All  that  we  know  of  0  in  any  of  these  formulas  is  that  its 
value  lies  between  0  and  + 1 . 

Cor.  2.  If,  on  applying  to  a  given  function  any  one  of  these 
completed  formulas,  the  last  term  becomes  0,  or  approaches  0 
as  its  limit,  as  n  increases,  the  formula  evidently  develops  this 
function  ;  if  not,  the  formula  fails. 

tjn      ?/        iftl~  ?y 

101.  Since  f-  =  -  •  — ;  and  since  -  is  very  small  when  y 

\n_      n     |  n  —  1  n 

vn 
is  finite,  and  n  is  very  large  ;   each  value  of  —  is  a  very  small 

[n 

part  of  its  preceding  value. 

vn 
Hence  j —  ==0,  token  y  is  finite  and  n  increases  indefinitely. 

Cok.  Jff"(x)  dues  not  become  infinite  with  n,  Taylor's  and 
Maclaurin's  formulas  give  the  true  development  of  f(x-fy)  and 
f(x)  respectively. 

102.  To  prove  that  Maclaurin's  formida  develops  ax. 
Here        fn  (x)  =  (log  a)  ■  a?  ;  .  • ./» {6  x)  =  (log  a)  ■  a**, 

i               /•»//)   \  as"      (xlog  a)"    fl~ 
and  /"  (0  as)  —  =  -i -2 — '—  aex. 

\n  \n 

Since  a6x  is  finite,  and  ^ — r= — i-  =  0  as  n  increases  indefi- 
nitely (§  101)  ;  ^ 

[n 

as  ?i  increases,  and  the  formula  develops  the  function  (§  100, 
Cor.  2). 


106  DEVELOPMENT    OF   FUNCTIONS   IN   SERIES. 

103.  To  prove  that  Ifaclaurin's  formula  develops  sinx  and 
cosx. 

The  nth  derivative  of  each  of  these  functions  is  finite,  how- 
ever great  n  may  be  ;  hence  Maclaurin's  formula  develops  both 
of  them  (§  101,  Cor.). 

104.  To  determine  for  ivliat  values  of  x  Maclaurin's  formula 
develops  log(l-\-x). 

The  formula  gives  (§  96), 

»_j£,«l_i£  . (  — l)"-ss— 1 

1       2        3        4  ri-1 

(  — 1)»-  1xn 


log(l  +  a0=4-±-  +  4--4-  +  ...+ 


The  ratio  of  the  ?ith  term  to  the  term  before  it  evidently 
approaches  —  x  as  n  increases.  Hence,  if  x  is  numerical^ 
greater  than  1 ,  the  series  is  divergent,  and  cannot  be  the  devel- 
opment of  log(l  +  #)-  We  need,  therefore,  to  examine  the 
value  of  the  last  term  of  the  formula  only  for  values  of  x  be- 
tween —  1  and  -f-1. 

(_l)n-ljw_1 

/"(*)  =  ■ 


J  V     V  n        \\  +  6x) 


i  —  lY'1 

For  values  of  x  between  0  and  + 1 ,  •* ' —  and  , 

n  \l  +  0x 

each  approaches  0  as  n  increases ;  hence  the  formula  develops 

log  (1  +  x)  for  these  values  of  x. 

"When  x  lies  between  0  and  —  1 ,  let  xx  represent  its  absolute 
value  ;  then  x  =  —  a^,  and  log(l  -f-  x)  =  log(l  —  xx) . 

Using  the  second  form  of  the  formula,  we  have,  numerically, 

f»(QX )  a-or-w  =  (i-o)*-w 

J    K     V       \n-\  (l-dx,)'1 


l  —  exyj     1—8X1 


taylor's  and  maclaurin's  formulas.        107 


For  values  of  xv  between  0  and  +1,  ! —  is  finite,  and 

1  —  6xx 

A*.    A™  \n-l 

f-i — —J]     approaches  0  as  n  increases.     The  formula  there- 

\i-exj 

fore  develops  log  (l  +  x)  when  x  lies  between  0  and  —  1. 

Hence,  Maclaurin's  formula  develops  loy(\-\-  x)  for  values  of 
x  between  —  1  and  +  1. 

105.    To  determine  for  what  values  of  x  Maclaurin's  formula 
develops  (l  +  x)m. 
The  formula  gives 

(1  +  X)m  =  1  +  mx  +  mi™-1)^  + ... 

m(m— l)  —  (m  — n+1)    _ 

H ■ ,{B  • 

If  m  is  a  positive  whole  number,  fn(0x)  —  =0  when  »  =  wi-f-1 ; 

In 

hence,  in  this  case,  Maclaurin's  formula  develops  (l+#)m  in  a 

finite  series  of  m  + 1  terms. 

If  m  is  negative  or  fractional,  the  series  is  infinite.    The  ratio 

of  the  (n  +  l)th  term  to  the  ?ith  is —  aj,  which  approaches 

n 

—  x  as  n  increases.  The  series  therefore  is  divergent,  and  can- 
not equal  (1  +  x)m  when  x  is  numerically  greater  than  1.  Hence 
we  need  examine  the  value  of  the  last  term  of  the  formula  only 
for  values  of  x  between  + 1  and  —  1 . 

Here       /»(■)  =  ™0»-l)"(™-'H-l) 


m(m  — l)"-(m  —  w-t-1)   »" 


When  #  lies  between  0  and  1,  and  ?i  >  m, 

1 


1 


(1  +  0*)"- 


■<1. 
(1+fla?)"- 


108  DEVELOPMENT   OF   FUNCTIONS   IN   SERIES. 

An  increase  of  1  in  the  number  of  terms  multiplies  the  factor 

in  brackets  by  x,  which  approaches  —  x  as  n  increases. 

n  +  1 

Hence  the  last  term  of  the  formula  approaches  0  as  n  increases  ; 
and  the  formula  develops  (l  +  x)m  for  values  of  x  between  0 
and  +1. 

Using  the  second  form  of  the  formula,  we  have 

piex)^-6^1*'1 

n  —  1 


m(m-l)  — (m-n+1)     " 

In  — 1 


i—oy(i+ox)' 


1  +  VxJ       1— ( 


For  values  of  x  between  0  and    —  1 ,    -^^t — lJ_    is   finite  : 

1-6 

i a  \» 

approaches  0  as  n  increases  ;  and  an  increase  of  1  in 


\l  +  0xj 


the  number  of  terms  multiplies  the  factor  in  brackets  by x. 

n 
which  approaches  —  x  as  n  increases.     Hence  Maclaurin's  for- 
mula develops  (l  +  x)m  for  all  values  of  x  between  —1  and  +1. 


106.   The  Binomial  Theorem.    Since  (a  +  x)m  =  am(  l-f 

V     a 

(x\m 
1  +  -)   can  be  developed  by  Maclaurin's  formula,  when  x 

is  numerically  less  than  a  (§  105)  ;  therefore,  in  this  case, 

(a  +  x)m  =  am  +  mam-lx  +  m{m~1)  a—*aS+  •••  (1) 

For  like  reason,  when  a;  is  numerically  greater  than  a, 

(x  +  a) "  =  xm  + mxm~la  +  m(7^~1). xm-aa*+—  (2 ) 

Hence  one,  and  only  one,  form  of  the  development  of  (a  +  x)m 
holds  for  anv  set  of  values  of  a  and  x. 


THE   COMPUTATION   OF   77\  109 

107.    To  develop  tan *x,  and  find  the  value  of  tt. 

tan  ~1x=  (  — — •• 
J  1  +  ar 

When  a;  lies  between  —  1  and  + 1 , 

— i-j  =  (1+s8)-1  =  1-  aj»+  x*-  o:c4-  xs ;  §  105. 

1  +  xr 

.*.  | ;  =  I  dx  —  I  ardx  +  J  x'cto  —  |  a;6dc 

+  faftfa; 

Hence,  if  x  is  numerically  less  than  1, 

/yO  /%*5  /y,t  /y»9 

tan-1a?  =  a?-  —  +  —  -  —  +  - (1) 

3        5        7        9  K  J 

C  =  0,  since  tan_1x  =  0  when  x  =  0. 

If  we  put  a;  =  \|r,  equation  (1)  becomes 


tan 


,     1      tt         1A      1,1         1  \ 

\3  =  6=\3V1-9+i5-T^  +  "V 


.•.*r  =  2V3[l--i-f- —  +  —^  =  8.141592-+. 

^       9      45      189  J 

108.    To  develop)  sm_1x,  and  find  the  value  o/tt. 

.   _i             ,     a*3     .    1  •  3  x-5   .    1  •  3  •  5  x7    , 
sm  Lx  =  x-i , 

2- 32-4- 52- 4-6- 7 

when  x  lies  between  —  1  and  -f- 1  • 

A=_I(i+J.+JL+JL  + 

6      2\       24      640      7168 

or,  77  =  3.141592  +  . 

It  was  by  means  of  this  series  that  Sir  Isaac  Newton  com- 
puted the  value  of  tt. 


110 


DEVELOPMENT   OF   FUNCTIONS    IN   SERIES. 


109.  To  prove  geometrically  that  f  (a  +  h)  =  f  (a,)  +  h  f  (a-f-0h) , 
u        f(x)  andf(x)  being  continuous  and  finite  be- 
tween x  =a  and  x  =  a  +  h. 

Let  pp'p"  be  the  locus  of  y  =/(#) ,  a  =  oa, 
and  h  =  an  ;  then  /  (a)  =  ap  =  nm,  and 
/(a  +  7i)  =  Np".  The  curve  must  be  parallel 
to  the  chord  pp"  at  some  point  p'  between  p 
and  p".  Now,  ob  =  oa  +  ab  =  a  +  07t,  6  hav- 
ing some  value  between  0  and  1  ; 

hence  f'(a  +  #'0  =  tan  bdp'=  tan  mpp". 

.'.Mp"=PMtanMPp"=  hf'(a  +  0K)  ; 

.-./(a  +  h)  =  nm  +  Mp"=/(a)  +  hf'(a  +  $h). 


Examples. 


1.    Develop  (a2 +  ba?)i. 


A         /  ■>  ,  7   ->\i  ,  bar      b2x*  .    bsxe 

Ans.    {a-  +  &ar)  s  =  a  + — ■  +  — — 

2  a      8  a3      16  a5 


2 .    Prove  that  tan  x  =  x  -\ 1 1- 

3        15 


3.    Prove  that  sec  x  =  1 H 1 1 1- 

2        24        720 


4.  Given  f(x)  =  2  a?  —  3  a? -\-  ix  —  3,   to   find   the  value   of 
fix  +  h) ,  h  being  a  variable  increment  of  x. 

Here  f(x)  of  Taylor's  formula  is 

2a?  —  3x2  +  4.t  —  3,  and  y  =  h; 

.-.  f'(x)  =  6x2-Gx  +  4,    /"(«)=  12 x  —  6, 

/'"(«)  =  12,  and/'v(a-)  =  0. 

.*./(*  +  /*)  =  2^  —  3a,-2  +  4a;  —  3  +  (Gar2  —  6a;  +  4)  h 
+  (Gx-3)h2  +  2h3. 

5 .  Given  /(»)«=  2  as5  —  3  a?,  to  find  f(x  +  h). 


EXAMPLES.  Ill 

6.  Develop  sin  (x  +  y) . 

sin  (x  +  y)  =  sin  as  f  1  —  iC  4.  Jl.  __  JL  j A 

=  sin  a  cos  y  +  cos  a;  sin  y.  §§93,94. 

7.  Prove  that  cos  (x  +  y)  =  cos  a;  cos  y  —  sin  x  sin  y. 

8.  Prove  that  ecosx  =  e(l-  —  +  —  - ^1^  +  ...\ 

V    12.    LL     Li       ; 

9.  Prove  that  los  ( 1  +  sin  a?)  =  a?  —  —  +  —  —  —  +  ... 

10.  Develop  e8ini.       eBinx  =  l  +  x+ ~  —  ~4-  —  —— ±  ... 

11       H+|l       H  + 

11.  Develop 


V&"  —  c-x2 


12.   Develop  a^ex.  arJex  =  ar!  +  arJ  +  — +  — + 

11     11 


CHAPTER  VIII. 


MAXIMA    AND    MINIMA. 


110.  A  maximum  value  of  a  function  of  a  single  variable  it> 
a  value  that  is  greater  than  its  immediately  preceding  and  suc- 
ceeding values.  A  minimum  value  is  one  that  is  less  than  its 
immediately  preceding  and  succeeding  values. 

Therefore,  if  we  conceive  x  as  alwaj-s  increasing,  f(x)  must 
be  an  increasing  function  immediately  before,  and  a  decreasing 
function  immediately  after,  a  maximum  ;  also,  immediately  be- 
fore a  minimum,  f(x)  is  a  decreasing,  and  immediately  after 
an  increasing  function. 

Hence,  f  (x)  is  2>ositive  before  and  negative  after  a  maximum 
o/f(x),  and  negative  before  and  positive  after  a  minimum. 

111.  From  §  110.  f'(x)  must  change  its  sign  as  f(x)  passes 
through  either  a  maximum  or  a  minimum.  But,  to  change  its 
sign,  f'{x)  must  pass  through  0  or  oo. 

Hence,  any  value  of  x  that  renders  f  (x)  a  maximum  or  a 
minimum  is  a  root  of  f'(x)  =  0  or  f'(x)  =  oo. 

The  converse  of  this  theorem  is  not  true  ;  that  is,  any  root 
of  /'(#)  =  0  or  oo  does  not  necessarily  render  f(x)  a  maximum 
or  a  minimum.  These  roots  are  simply  the  critical  values  of  x, 
for  each  of  which  the  function  is  to  be  examined. 

To  illustrate  geometrically  the  preceding  definitions  and 
Y  principles,  suppose  a'h'  to  be  the  locus 

of  y  =  f[x).  Then,  by  definition,  aa', 
cc',  and  ee'  are  maxima,  and  bb'  and  dd' 
are  minima  of  f(x) .  In  passing  along 
the  curve  from  left  to  right,  the  slope 
of  the  curve  f'(x)  is  positive  before, 
and  negative  after,  a  maximum  ordi- 
nate ;  and  negative  before,  and  positive  after,  a  minimum 
ordinate. 


METHODS   OF   FINDING    MAXIMA   AND   MINIMA.       113 

Moreover,  at  a  point  whose  ordinate  f(x)  is  a  maximum  or 
a  minimum,  the  curve  is  either  parallel  or  perpendicular  to  the 
axis  of  x,  and  therefore  f'{x)  =  0  or  oo. 

At  x  =  oh,  /'(ce)  =  0,  but  hh'  is  neither  a  maximum  nor  a 
minimum  of  f(x) . 

112.  Whether  any  one  of  the  critical  values  of  x  renders /(.t) 
a  maximum  or  a  minimum  can  be  determined  by  one  of  the 
following  methods  : 

First  Method.  In  this  method  we  determine  directly  whether 
f'(x)  changes  from  positive  to  negative,  or  from  negative  to 
positive,  as  x  passes  through  a  critical  value. 

Let  a  be  a  critical  value,  and  h  a  very  small  quantity.  In 
f'(x)  substitute  a  —  h  and  a  -f  h  for  x. 

If  f'(a  —  h)  is  positive,  and  f'(a  +  h)  negative, 

/(a)  is  a  maximum.  §  110 

If  f'(a-h)  is  negative,  and  /'(a  +  /i)  positive, 
/(a)  is  a  minimum. 

If  f'(a  —  h)  and  f'(a  -\-  h)  have  the  same  sign, 
/(«)  is  neither  a  maximum  nor  a  minimum. 

Second  Method.  This  method  applies  only  to  the  roots  of 
f'(x)  =  0.  Let  a  be  a  cntical  value  of  x.  Developing  /(as— h) 
and  f(x  -f  h)  by  Taylor's  formula,  substituting  a  for  x,  trans- 
posing /(«),  and  remembering  that  /'(«)  =  0,  we  have 

/(a-/0-/(a)=/"(a)^-/"'(«)|+/-(a)|-...,     (1) 

and     /(a+A)-/(a)=/''(a)|V/''(a)|8+/nr(a)1|V--.-      (2) 

If  h  be  taken  very  small,  the  sign  of  the  second  member  of 
either  (1)  or  (2)  will  be  the  same  as  the  sign  of  its  first  term. 
Hence,  if  /"(«)  is  negative,  f(a)  is  greater  than  both /(a  —  It ) 
and  /(<x+7i),  and  therefore  a  maximum;  while,  if  /"(a)  is 
positive,  /(a)  is  less  than  both  f(a  —  h)  and  /(a  -f-  h) ,  and 


114  MAXIMA   AND   MINIMA. 

therefore  a  minimum.  If  f"(a)  is  0,  and  /'"(a)  is  not  0, 
/(a)  is  neither  greater  than  both  /(a  —  h)  and  f(a  +  h) ,  nor 
less  than  both,  and  is  therefore  neither  a  maximum  nor  a  mini- 
mum. If /'"(a),  as  well  as /"(a),  is  0,  and/iv(a)  is  negative, 
/(«)  is  greater  than  both  /(a  —  h)  and  f{a  +  /i),  and  therefore 
a  maximum;  while,  if  flY(a)  is  positive,  j\a)  is  a  minimum, 
and  so  on. 

Hence,  if  a  is  a  enticed  value  obtained  from  f'(x)  =  0,  substi- 
tute a  for  x  in  the  successive  derivatives  of  f(x).  If  the  first 
derivative  that  does  not  reduce  to  0  is  of  an  odd  order,  f  (a)  is 
neither  a  maximum  nor  a  minimum  ;  but,  if  the  first  derivative 
that  does  not  reduce  to  0  is  of  an  even  order,  f  (a)  is  a  maximum 
or  a  minimum,  according  as  this  derivative  is  negative  or  positive. 

113.  Maxima  and  minima  occur  alternately. 

Suppose  that  a  <  b,  and  that  /(a)  and  f(b)  are  maxima 
of  f(x).  When  x  =  a  -f  h,  f{x)  is  decreasing;  and,  when 
x=b  —  h,  f(x)  is  increasing,  h  being  very  small.  But,  in 
passing  from  a  decreasing  to  an  increasing  state,  f(x)  must  pass 
through  a  minimum.  Hence,  between  two  maxima,  there  must 
be  at  least  one  minimum. 

In  like  manner,  it  can  be  proved  that  between  two  minima 
there  must  be  at  least  one  maximum. 

114.  The  solution  of  problems  in  maxima  and  minima  is 
sometimes  facilitated  by  the  following  considerations  : 

(a)  Any  value  of  x  that  renders  c  -/(x)  a  maximum  or  a  minimum,  c 
being  positive,  renders  f(x)  a  maximum  or  a  minimum. 

(b)  Any  value  of  x  that  renders  log0/(.r)  a  maximum  or  a  minimum, 
renders  f(x)  a  maximum  or  a  minimum,  a  being  greater  than  unity. 

(c)  Any  value  of  x  that  renders /(x)  a  maximum  or  a  minimum,  renders 
a  minimum  or  a  maximum. 

/GO 

(d)  Any  value  of  x  that  renders  c  +f(x)  a  maximum  or  a  minimum, 
renders  f(x)  a  maximum  or  a  minimum. 

(e)  Any  value  of  x  that  renders  f(x)  positive,  and  a  maximum  or  a 
minimum,  renders  [/(x)]n  a  maximum  or  a  minimum,  n  being  a  positive 
whole  number. 


EXAMPLES.  115 


Examples. 


1.  Find  what  values  of  x  render  4ar3  —  15. t2  +  12  a;  —  1   a 
maximum  or  a  minimum. 

Here         f(x)  =  4aj8—  15aj2  +  12a?—  1; 

,'.f{x)  =  12.it  -  30.r  -f  12,  and  f"(x)  =  24  a;  -  30  ; 

hence  the  roots  off'(x)  =  12ar  —  30&  +  12  =  0,  which  are  $  and 
2,  are  the  critical  values  of  a;  (§  111). 

But  /"(*)  =  [24a; -30] 4  =  -18, 

and  /"(2)  =  [24a;-30]2  =  -f  18; 

hence  when  x  =  -|-,   the  function  is  a  maximum  (§  112),   and 
when  x  =  2,  it  is  a  minimum. 

2.  Find  the  maxima  and  minima  of  ar3  —  9  ar  +  15  a;  —  3. 

Here        f'(x)  =  3 ar  -  18a;  +  15,  and  f"(x)  =  G x  -  18  ; 

therefore    5  and   1,   the  roots  of  f'(x)  =  Sx2—  18a;  +  15  =  0, 
are  the  critical  values  (§  111). 

/"(5)  =  [6aj-18]5  =  +12,  and/"(l)  =  -12; 

.-./(5),  or  [ar3 -9ar+15.r-3]5,  [=-28]  isamin., 

and  /(l),  or  [ar3  — 9ar +15 a;  — 3]^  [=  4]   is  a  max. 

Let  the  student  construct  the  locus  of  y  =  ar3— 9ar2+  15a;  —  3. 
and  thus  exhibit  these  results  geometrically. 

3.  Examine  ar3  —  3  ar+  3  a;  +  7  for  maxima  and  minima. 
Here         /'(*)  =  3a,-2- 6.t +3,  f"(x)  =Gx-G,  and/"'(aj)=6  ; 

therefore  1  is  the  critical  value. 

But  /"(l)  =  [6a;-6]1  =  0,  and  /"'(1)  =  6  ; 

hence  the  function  has  neither  a  maximum  nor  a  minimum  (§  112). 

4.  Examine  x5  —  5x*-\-5xi— 1  for  maxima  and  minima. 
Here        f(x)  =  ar5  —  5  a;4  +  5  ar3  —  1 ; 


116  MAXIMA  AND  MINIMA. 

.'.f'(x)  =  5xi-20xi  +  loxi,  f '" (x)  =  20a8— 60 x2 +30 a, 
and  f"{x)=BOx*  —  120a +'80; 

therefore  1,3,  and  0  are  the  critical  values. 

Since       /"(1)  =  -  10,  /(1)[  =  0]  is  a  maximum. 
Since       /"(3)  =  +  GO,  /(3)  [  =  -28]  is  a  minimum. 
Since       /"(0)  =  0,  and /'"(0)  =  30, 

/(0)  is  neither  a  maximum  nor  a  minimum. 

5.  Examine  (x  —  l)4  (x  +  2)3  for  maxima  and  minima. 
Here        /'(«)  =  (a -  1)8(»  +  2)?(7a>  +  5), 

and  the  critical  values  are  —  2,  —  f ,  and  +  1. 

In  this  example,  the  first  method  is  to  be  preferred.     By  in 
spection,  we  see  that 

f'(—2  —  h)  and  /'(—  2  +  h)  are  both  positive  ; 
hence  /( —  2)  is  neither  a  maximum  nor  a  minimum  (§112) 

/'(_*_ft)is+,  and  f{-4  +  A)  is-; 
hence  /(~t)  *s  a  maximum  (§  112). 

/'(I  -  /t)  is  -,  and  /'(l  +  ft)  is  +  ; 
hence  /(l)  [=0]  is  a  minimum. 

6.  Examine  b-\-c(x  —  a)s  for  maxima  and  minima. 

Here        f(x)  =        2°        ; 
3(#  — a)* 

and  the  critical  value  is  a,  the  root  of  f'(x)  =  a>. 

2c  .  ,  2c  .      , 

-  is  — ,  and —  is  +  ; 


3  (a  —  h  +  a)1*  (a  +  ft  +  a)* 

.'./(a)  [=6]  is  a  minimum. 

(ft    ,        /v\  3  ^      ^ 

7.    Examine  ^ — —  for  maxima  and  minima. 

a  —  2x 

tt  /»//  \       (a  —  x)2(4x  —  a) 

Here       /'  (a?)  =  ^ —        \       — *-  ; 
(a  —  2  a;)- 


EXAMPLES.  117 

and  the  equations  f'(x)  =  0  and  f'(x)  =  oo  give  -,  -,  and  a  as 

4   2 
the  critical  values. 

By  inspection,  we  see  that  f'(x)   changes  from  negative  to 

positive  when  a;  =  - ;   hence/f  -  J  is  a  minimum.     But,  as/' (x) 

does  not  change  its  sign  when  x  =  a  or  -,  f(a)  and/[-jare 
neither  maxima  nor  minima. 


8.    Examine  c  +  V4crar  —  2  ax3  for  maxima  and  minima. 
By  (d),  (e),  and  (a)  of  §  114,  any  value  of  x  that  renders 
c  +  a/4 a?ar  —  2  ax3  a  maximum  or  a  minimum,  renders 


V4  crar  —  2  (far5,  4  e^ar2  —  2  aar3,  and  2  aar  —  ar3 
a  maximum  or  a  minimum. 

Hence,  let  f(x)  =  2  ax2  —  ar3,  etc. 


,d?is.  When  x  =  0,  c+  v4«2ar  —  2 aar3  is  a  minimum  ; 
"     #  =  |a,       "  "         is  a  maximum. 

9.  "What  values  of  x  render  2a;3  —  21  x2  +  36 x  —  20  a  maxi- 
mum or  a  minimum  ? 

Ans.  /(I)  is  a  max.  ;  /(6)  is  a  min. 

10.  Examine  Sx5  —  125  a,"3  +  21G0a;  for  maxima  and  minima. 
Ans.  /(— 4)  and/(3)  are  max.  ;  /(—  3)  aud/(4)  are  min. 

1 1 .  Examine  Xs  —  3x*  +  Qx-\-7  for  maxima  and  minima. 

Ans.  It  has  neither  a  max.  nor  a  min. 

12.  If /'(a-)  =  ar3(a;-l)2(a;-2)3(ar-3)4,  what  values  of  x 
render  fix)  a  maximum  or  a  minimum? 

Ans.  /(0)  is  a  max.  ;  /(2)  is  a  min. 

13.  Examine  a;(a;  -J-  a)2(a  —  a;)3  for  maxima  and  minima. 

Ans.  /(—a)  and/(  - )  are  max.  ;  /[ J  is  a  min. 


118  MAXIMA   AND   MINIMA. 

14.  Examine for  maxima  and  minima. 

x  -10 

.4ws.  /(4)  is  a  max. ;  /(16)  is  a  min. 

(x  -f-  2Y 

15.  Examine  *-  „  for  maxima  and  minima. 

(x-sy 

Ans.  /(3)  is  a  max.  ;  /(13)  is  a  min. 

16.  Prove  that  sin  x  +  cos  a;  is  a  maximum  when  x  =  -- 

4 


17.  Examine  — - —  for  maxima  and  minima. 


log  a; 

Ans. 

loge 


Ans. is  a  min. 


18.  Prove  that  of  is  a  maximum  when  x  =  e. 

19.  Prove  that  sin  #(1+  cos  x)  is  a  maximum  when  x  =  -• 

o 

20.  Prove  that is  a  maximum  when  x  =  cos  x. 

1  +  ajtanic 

21.  Examine  the  curve  y  =  x?  —  Sx2  —  24  a;  +  85  for  maxima 

and  minima  ordinates. 

Ans.  113  is  a  max. ;  5  is  a  min. 

22.  Examine  y  =  x3  —  9a^  +  24 #+16  for  maxima  and  min- 
ima ordinates. 

Ans.  36  is  a  max. ;  32  is  a  min. 

23.  Examine  y  =  x3  —  Sx2  —  9x-\-5  for  maxima  and  minima 

ordinates. 

Ans.  10  is  a  max.  ;   —  22  is  a  min. 

24.  Examine  y  =  x5  —  5  cc4  +  5  ar5  +  1  for  maxima  and  minima 

ordinates. 

Ans.  2  is  a  max. ;   —26  is  a  min. 

25.  Examine  y=  sin3#cosa  for  maxima  and  minima  ordi- 
nates. 

Ans.  When  x  =  -Jtt,  y  =  ^-V3,  a  max. 


GEOMETKIC    PROBLEMS. 


119 


Geometric  Problems. 

1.    Find  the  altitude  of  the  maximum  cylinder  that  can  be 
inscribed  in  a  given  right  cone. 

Let  ik  be  the  cylinder  inscribed  in  the  d 

given  cone  dab.  Let  a=DC,  &=ac,  ?/=mc, 
x  =  im,  and  V=  the  volume  of  the  cylinder  ; 
then  V=  Tvxy2. 

From  the  similar  triangles  adc  and  idh, 
we  find 

y  =  -(a-x); 
a 


V=  7T—  x  (a  —  x)2, 


Fig.  28. 


Fig.  29. 


which  is  the  function  whose  maximum  is  required. 

Let  f{x)  =  x  (a  -  x) 2,  etc.  §  1 14,  (a) . 

Ans.  The  altitude  of  the  cylinder  =  £  that  of  the  cone. 

2.  Find   the   altitude   of   the   maximum   cone   that   can  be 
inscribed  in  a  sphere  whose  radius  is  r. 

Let  acd  and  acb  be  the  semicircle  and 
the  triangle  which  generate  the  sphere  and 
the  cone.  Let  x  =  ab,  y  =  bc,  and  V= 
the  volume  of  the  cone  ;  then  V=  \trxy2. 

Since       y2  =  AB'BD  =  <c(2r—  a;), 

V=  ^7tx,2(2  r  —  x) , 

which  is  the  function  whose  maximum  is  required. 

Ans.  The  altitude  of  the  cone  =  f  the  radius  of  the  sphere. 

3.  Find  the  altitude  of  the  maximum  cylinder  that  can  be 
inscribed  in  a  sphere  whose  radius  is  r. 

Let  x  =  ab,  and  y  =  be  ; 

then  V=2irxy2=2-n-x{i- —  x2). 

Ans.  Altitude  =  f?'V3.  Fig?3o. 


120  MAXIMA  AND   MINIMA. 

4.  Find  the  maximum  rectangle  that  can  be  inscribed  in  an 
ellipse  whose  semi-axes  ai*e  a  and  b. 

Ans.  The  sides  are  aV2  and  &V2  ;  the  area  =  2ab. 

5.  Find  the  maximum  cylinder  that  can  be  inscribed  in  an 
oblate  spheroid  whose  semi-axes  are  a  and  b. 

A)is.  The  radius  of  the  base  =  ^aV6  ;  the  altitude  =  f  &V3. 

6.  The  capacity  of  a  closed  cylindrical  vessel  being  c,  a  con- 
stant, what  is  the  ratio  of  its  altitude  to  the  diameter  of  its 
base,  when  its  entire  inner  surface  is  a  minimum?  What  is  its 
altitude  ? 

Let  y  equal  the  radius  of  the  base,  x  the  altitude,  and  S  the 
entire  inner  surface  ;  then 

c  =  irxy2,  (1) 

and  #  =  2  Try2  +  2  wyx.  (2) 

From  (1), 

ty  =  -JL.  (S) 

dx         2x  V  ' 

From  (2), 

^  =  4^  +  2^  +  2^.  (4) 

dx  dx  dx 

Since  — ■  =  0  when  S  is  a  minimum,  from  (4)  we  have 
dx 

dy  =         y     , 

dx         2  y  +  x 

From  (3)  and  (5), 


(5) 


_J/_  = y~,ovx  =  2y.  (6) 

2x      2y  +  x  y  v  } 

Hence,  as  S  evidently  has  a  minimum  value,  it  is  a  minimum 
when  the  altitude  of  the  cylinder  is  equal  to  the  diameter  of  its 
base. 

From  (1)  and  (6), 


•\27T 


GEOMETRIC    PROBLEMS.  121 

This  problem  might  have  been  solved  like  those  preceding  it ; 
that  is,  by  eliminating  y  between  (1)  and  (2)  at  first.  In  many 
problems,  however,  the  method  given  in  this  example  is  much 
to  be  preferred. 

7.  The  capacity  of  a  cylindrical  vessel  with  open  top  being 
constant,  what  is  the  ratio  of  its  altitude  to  the  radius  of  its 
base  when  its  inner  surface  is  a  minimum  ? 

Ans.  Its  altitude  =  the  radius  of  its  base. 

8.  A  square  piece  of  sheet  lead  has  a  square  cut  out  at  each 
corner  ;  find  the  side  of  the  square  cut  out  when  the  remainder 
of  the  sheet  will  form  a  vessel  of  maximum  capachy. 

Ans.  A  side  =  £  the  edge  of  the  sheet  of  lead. 

9.  Find  the  arc  of  the  sector  that  must  be  cut  from  a  circular 
piece  of  paper,  that  the  remaining  sector  may  form  the  convex 
surface  of  a  cone  of  maximum  volume,  r  being  the  radius  of  the 

cirele"  Ans.  The  arc  =  2  itr (1  -  |V6) . 

10.  A  person,  being  in  a  boat  3  miles  from  the  nearest  point 
of  the  beach,  wishes  to  reach  in  the  shortest  time  a  place  5  miles 
from  that  point  along  the  shore  ;  supposing  he  can  walk  5  miles 
an  hour,  but  row  only  at  the  rate  of  4  miles  an  hour,  required 
the  place  where  he  must  land. 

Ans.  1  mile  from  the  place  to  be  reached. 

11.  Find  the  maximum  right  cone  that  can  be  inscribed  in  a 
given  right  cone,  the  vertex  of  the  required  cone  being  at  the 
centre  of  the  base  of  the  given  cone. 

Ans.  The  ratio  of  their  altitudes  is  £. 

12.  A  Norman  window  consists  of  a  rectangle  surmounted  by 
a  semicircle.  Given  the  perimeter,  required  the  height  and  the 
breadth  of  the  window  when  the  quantity  of  light  admitted  is  a 
maximum. 

Ans.  The  radius  of  the  semicircle  =  the  height  of  the  rectangle. 


122 


MAXIMA   AND   MINIMA. 


13.  Prove  that,  of  all  circular  sectors  having  the  same 
perimeter,  the  sector  of  maximum  area  is  that  iu  which  the  cir- 
cular arc  is  double  the  radius. 

14.  Find  the  maximum  convex  surface  of  a  cylinder  inscribed 
in  a  cone  whose  altitude  is  6,  and  the  radius  of  whose  base  is  a. 

Ans.  Maximum  surface  =  %-rrab. 

15.  Find  the  altitude  of  the  cylinder  of  maximum  convex 
surface  that  can  be  inscribed  in  a  given  sphere  whose  radius 
is  r.  Ans.  Altitude  =  rs/%. 

16.  Find  the  altitude  of  the  cone  of  maximum  convex  surface 
that  can  be  inscribed  in  a  given  sphere  whose  radius  is  r. 

Ans.  Altitude  =  f  r. 

17.  Find  the  altitude  of  the  parabola  of  maximum  area  that 

can   be    cut   from   a  given  right  circular 
cone. 

Let  ob  =  2  6,  oc  =  a,  and  bm  =  x ;  then 

qq'  =  2  mq  =  2  Vmb  •  MO 


=  2Vx(2b-x), 
Also,    BO  :  bm  :  :  oc  :  MP, 
or  2  b  :  x  : :  a  :  mp  : 


".  area  =  f 


MP 


2a 


ax 
Tb 


mp  =  —  ~\/xJ(2b 
3b        v 


x), 


§  66,  Ex.  1. 
§  114. 


Let  f(x)  =  x5(2  b  —  x) ,  etc. 

Ans.  The  parabola  is  a  maximum  when  its  altitude  mp  is  f 
the  slant  height  of  the  cone. 

18.    What  is  the  altitude  of  the  maximum  cylinder  that  can 
be  inscribed  in  a  given  prolate  spheroid  ;  that  is,  in  a  solid  gen- 
erated by  the  revolution  of  a  given  ellipse  about  its  major  axis  ? 
Ans.  Altitude  =  the  major  axis  divided  by  V3. 


GEOMETRIC   PROBLEMS.  123 

19.  Find  the  number  of  equal  parts  into  which  a  given  num- 
ber a  must  be  divided  that  their  continued  product  may  be  a 

maximum.       ,        _.  ,         _  a 

Ans.   I  he  number  of  parts  =  -,  and  each  part  =  e. 

20.  A  privateer  has  to  pass  between  two  lights  a  and  b,  on 
opposite  headlands.  The  intensity  of  each  light  is  known,  and 
also  the  distance  between  them.  At  what  point  must  the  pri- 
vateer cross  the  line  joining  the  lights,  so  as  to  be  in  the  light 
as  little  as  possible  ? 

Let  d  =  the  distance  ab,  and  x  the  distance  from  a  of  any 

point  p  on  ab.     Let  a  and  b  be  the  intensities  of  the  lights  a 

and  b  respectively,  at  a  unit's  distance.      By  a  principle  of 

Optics,  the  intensity  of  a  light  at  any  point  equals  its  intensity 

at  a  unit's  distance  divided  b}'  the  square  of  the  distance  of  the 

point  from  the  light. 

a  b  ' 

Hence  —7  + —  is  the  function  whose  minimum  we  seek. 

xr       (d  —  x)- 

Ans.  x  =  — • 

as  -f-  b* 

21.  The  flame  of  a  lamp  is  directly  over  the  centre  of  a  cir- 
cle whose  radius  is  r ;  what  is  the  distance  of  the  flame  above 
the  centre  when  the  circumference  is  illuminated  as  much  as 
possible  ? 

Let  a  be  the  flame,  p  any  point  on  the  circum- 
ference, and  £  =  ac.  By  a  principle  of  Optics, 
the  intensity  of  illumination  at  p  varies  directly 
a.s  sin  cpa,  and  inversely  as  the  square  of  pa. 

Hence ■ is  the  function  whose  maxi- 

C^+f)1  .  Fig.  32. 

mum  is  required,  in  which  r  is  the  radius  of  the 

circle,  and  a  is  the  intensity  of  illumination  at  a  unit's  distance 

from  the  flame.  Ang    x=s^r^ 

22.  On  the  line  joining  the  centres  of  two  spheres,  find 
the  point  from  which  the  maximum  of  spherical  surface  is 
visible. 


124 


MAXIMA   AND   MINIMA. 


Let  cp  =  r,  cp  =  R,  cc  =  r7,  and  ca  =  as,  a  being  any  point  on 
wim.    From  A  draw  the  tangents  Ap  and  ap  ; .  then  the  sum  of  the 

zones  whose  altitudes  are  nm 
and  nm  is  the  function  whose 
maximum  is  required. 


Since  en  =  —  ,  by  Geometry 
Fis- 33-  we  have 

zone  nm  =  2  -n-r  •  nm  =  2  irr(r  —  en)  =  2  ■ 


Hence  2tt 


r2+B2- 


imum  is  sought. 


?*' 


sr- 


d  —  x 


is  the  function  whose  max- 
Ans.  x- 


rl+ffi 

23.  Assuming  that  the  work  of  driving  a  steamer  through 
the  water  varies  as  the  cube  of  her  speed,  find  her  most  eco- 
nomical rate  per  hour  against  a  current  running  c  miles  per 
hour. 

Let  v  =  the  speed  of  the  steamer  in  miles  per  hour. 

Then        avs  =  the  work  per  hour,  a  being  constant ; 
and  v  —  c  =  the  actual  distance  advanced  per  hour. 


Hence 


av° 
V  —  c 


=  the  work  per  mile  of  actual  advance. 


Aiis.  v  =  £c. 


CHAPTER   IX. 

FUNCTIONS  OF  TWO  OR  MORE  VARIABLES,  AND  CHANGE  OF 
THE  INDEPENDENT  VARIABLE. 

115.  Functions  of  two  or  more  Variables.  Since  any  inde- 
pendent variable  is  some  arbitrary  function  of  t,  t  representing 
time,  a  function  of  any  number  of  independent  variables  may 
be  regarded  as  a  function  of  the  single  variable  t,  and  therefore 
differentiated  by  the  rules  already  established. 

f(x,  y),  read  "function  of  x  and  y,"  represents  any  function 
of  x  and  y  ;  as,  x*  -f-  xy2  -f  xy  and  sin(a;  +  y) . 

f(x,  y,  z),  read  "function  of  x,  y,  and  z,"  represents  any 
function  of  x,  y,  and  z. 

116.  A  Partial  Differential  of  a  function  of  two  or  more  vari- 
ables is  the  differential  obtained  on  the  hypothesis  that  only  one 
of  the  variables  changes. 

117.  A  Total  Differential  of  a  function  of  two  or  more  vari- 
ables is  the  differential  obtained  on  the  hypothesis  that  all  its 
variables  change. 

118.  A  Partial  Derivative  of  a  function  of  two  or  more  vari- 
ables is  the  ratio  of  the  partial  differential  of  the  function  to 
the  differential  of  the  variable  that  is  supposed  to  change. 

119.  A  Total  Derivative  of  a  function  of  two  or  more  vari- 
ables, of  which  only  one  is  independent,  is  the  ratio  of  the  total 
differential  of  the  function  to  the  differential  of  the  independent 
variable. 

If  u  =  f(x,  y),  the  partial  differentials  of  u  with  respect  to 

x  and  y  are  written  dxu  and  d it,  or  —  dx  and  — dy;    and  the 

dx  dy 


126  FUNCTIONS    OF   TWO   OR    MORE   VARIABLES. 

partial    derivatives,    or    differential    coefficients,    are    written 

dTu        t  d,.u  ,     du        ,  du 

-JL-  and  -£-,  or  simplv  —  and  — 
dx  dy  dx  dy 

120.  The  toted  differential  of  a  function  of  two  or  more  vari- 
ables is  equal  to  the  sum  of  its  partial  differentials. 

For,  if  u  =f(x,  y) ,  it  is  evident,  from  the  general  principles  of 
differentiation,  that  du  can  contain  only  such  terms  as  are  of  the 
first  degree  in  dx  and  dy. 

Hence      du  =  cf>(x,  y)dx  -\-  cf>i(x,  y)dy,  (1) 

in  which  cj>(x,  y)  and  <f>i(x,  y)  represent  the  sums  of  the  co- 
efficients of  dx  and  dy  in  the  different  terms  of  du. 

Let  x'  and  y'  be  any  set  of  values  of  x  and  y  ;  then  we  have 

du  =  <f>(x',  y')dx+  <£i(#\  y')dy.  (2) 

Let  y  be  regarded  as  constant ;  then  dy  =  0,  <£(#',  y')  remains 
unchanged,  and  (2)  becomes 

dxu  =  <f>(x',  y')dx.  (3) 

If  x  is  constant,  (2)  becomes 

dyu=cf>1(x',y')dy.  (4) 

Adding  (3)  and  (4),  and  remembering  that  x'  and  y'  are  any 
values  of  x  and  ?/,  we  have  in  general 

dxu  +  dyu  =  (f>(x,  y)dx  +  <£i(#,  y)dy  =  du. 

Since  a  similar  process  of  reasoning  could  be  extended  to  a 
function  of  n  variables,  the  theorem  is  proved. 

If  x  and  y  were  not  independent,  the  demonstration  given 
above  would  still  hold ;  for  the  idea  of  a  partial  differential  of  a 
function  sets  aside  any  question  concerning  the  dependence  of 
its  variables. 

121.  Signification  of  Partial  Derivatives.  From  §  31  it  is 
evident  that  a  partial  derivative  expresses  the  ratio  of  the  rate 


THE   TOTAL   DERIVATIVE.  127 

of  change  of  the  function  to  that  of  its  variable,  so  far  as  its 
rate  depends  on  the  variable  supposed  to  change  ;  and  that  a 
function  is  an  increasing  or  a  decreasing  function  of  any  one  of 
its  variables,  according  as  its  partial  derivative  with  respect  to 
that  variable  is  positive  or  negative. 

Examples. 

1 .  u  =  by-x  +  ex2  -j-  gy3  -f-  ex ;   find  du. 
Here        dxu  =  ( by2  4-  2  ex  -+-  e) dx, 

and  dyu  =  ( 2  byx  +  3  gy2)  dy  ; 

.-.  du  =  (by2  +  2  ex  +  e)  dx  -f-  ( 2  byx  +  3  gy2)  dy. 

2.  u  =  y*.  Ans.    du  =  yx  log  y  dx  -f-  xif^dy. 

y 

3 .  u  =  log  Xs.  du  =  -  dx  +  log  x  dy. 

y  ,        xdy  —  ydx 

4.  u  =  tan-1--  du  =  —  .  ,     , — 

x  x-  +  y 

5 .  u  —  y Bin ■.  du  =  y 8ln x  log y  cos  »  das  +  - — —  dy. 

«•'  cr  oc  •        covers  z     ^ 

,  ,as  ,  ydx  —  xdy 

6.  ?<  =  lo2tan  '-.  ait  = 


(ar  +  3T)tan  l- 

7.    «  =  -*  +  £•  ^-^'^  +  ||f?y. 

a2      6-  a-  6- 

122.  If  u  =f(x,  y,  z),  and  y=cf>(x),  and  z  =  <£1(#),  u  is 
directly  a  function  of  a*,  and  indirectly  a  function  of  a*  through 
y  ond  z.  If  u=f(z,  ?/),  and  y  =  <f>(x),  and  2  =  <fo(a;),  w  is, 
in  like  manner,  indirectly  a  function  of  x  through  y  and  z.  In 
all  such  cases  the  total  derivative  of  u  with  respect  to  x  can  be 
obtained  by  finding  the  value  of  u  in  terms  of  a,*,  and  differen- 
tiating the  result ;  but  in  many  cases  it  is  more  readily  obtained 
by  using  the  formulas  of  the  next  article. 


128  FUNCTIONS    OF   TWO    OR   MORE   VARIABLES. 

123.    If  u  =  f(x,y,z),  and  y  =  <f>(x),  and  z  =  <f>1(x), 
we  have       du  —~dx  -\-^-  dy  -\-^  dz  ',  §120 


7        du  7     .  du  7     .  du  7 

du  =  —  dx-\ dy  -\ dz  ; 

da;  fZy  dz 


~du~\  *_  du      du  dy      du  dz 
dx        dx      dy  dx      dz  dx' 


du     du         ,  du 


in  which  — ,    — ,  and  —  are  the  partial  derivatives,  and 
dx     dy  dz 

the  total  derivative  of  u. 


_dxj 


Cor.  1.    If  u=f(x,y),   and  y  =  $(x),  du  =  —dx  +  —dy; 

dx  dy 


du. 

dx 


_  du  ,  du  dy 
dx      dy  dx 


Cor.  2.    If  u—f(y,  z),  and  y=tj>(x),  and  z  =  <f>1(x), 
du 


du 

dx 


du  7     ,  du  j 
—  dy  +  —  dz; 

dy  dz 

_  cJm  dy  .  du  dz 

dy  dx      dz  dx 


du 


Cor.  3.   If  u  =  f(y),  and  y=cf>(x),  du  =  —dy\ 

dy 

du  _  du  dy 
' '  dx      dy  dx 

Rem.    To  make  the  above  theorem  and  corollaries  intelligible, 
the  signification  of  each  term  and  factor  must  be  had  clearly  in 


mind.     Thus,  in  the  theorem, 


du 
dx 


denotes  the  total  derivative 


of  u  with  respect  to  x,  that  is,  the  derivative  obtained  on  the 
hypothesis  that  x,  y,  and  z  are  all  changing  according  to  the 

given  conditions  ;  while  —  denotes  the  partial  derivative  of  u 
b  dx 

with  respect  to  x,  that  is,  the  derivative  obtained  on  the  hypo- 
thesis that  y  and  z  are  constant. 


*  Analysts  are  not  agreed  as  to  the  best  means  of  distinguishing  total 
from  partial  derivatives ;  but  we  shall  always  distinguish  the  total  deriva- 
tive of  a  function  of  two  or  more  variables  by  enclosing  it  in  brackets. 


EXAMPLES. 


129 


Examples. 


1 .    u  =  z2  +  if  +  zy-i  %  =  sin  .r,  and  y  =  ez;    find 


Here 


VZu" 
dx 


du  dy     du  dz 
dy  dx     dz  dx 


(i) 


du      o  o  ,        du      0 

rifz  i   dy       , 

—  =  cos.t,  and  -±  =  eF. 
dx  dx 

Substituting  these  values  in  (1),  we  have 

fC^}  =  (3y°-  +  z)e*  +  (2z  +  y)cosx 

=  (3e2x  +  sina,*)ex  +  (2sina,'4-ex)cosa; 
=  3  e3*  +  e*(sin  x  -f-  cos  a;)  -+-  sin  2  x. 

This  same  result  could  be  obtained  by  substituting  in  u  the 
values  of  y  and  z  in  terms  of  x,  and  then  differentiating. 


2.    u  =  tan-1  (#?/) ,  and  y  =  e*. 


Ans. 


+x) 


Vdu\  __  ex(l  + 
\_dx J      1  +  a,-2 

3.    Tt  =  e<tr(?/  —  z),  ?/  =  asina?,  and  2=cosa;. 


4.    ?t  =  tan    _,  and  x2  +  y2  =  r2. 
x 


5.    m  =  sin  -,  2  =  e1,  and  y  =  ar. 


6.   u  =  Vx2  +  ?/2,  and  ?/  =  mx  +  c. 


rtful  _  _  1        _         1 
\_dxj         y  01       Vr^a? 

—    =  (a:—  2)  —  cos— • 
[dx]      v  V        .c2 

["(ftf] _  (l  +  m°-)x 

\_dXJ         yffi  _|_  (mx 


130 


FUNCTIONS    OF   TWO   OK    MORE   VARIABLES. 


7.    u  =  sirr1^—  z),  y=3x,  and  2  =  4a3. 


du 

dx 


Vi-ar2 


9    x*y 

u  =  x*y- -y-  +  x\  and  y  =  log x. 


du 
dx 


=  ^[4(log^)2  +  3i], 


124.  Implicit  Functions.  The  equation  f(x,  y)  =  0  indicates 
that  either  y  or  x  is  an,  implicit  function  of  the  other.  It 
represents  any  equation  containing  x  and  y  when  all  its  terms 
are  in  the  first  member.  Implicit  functions  have  been  differen- 
tiated heretofore,  but  the  formula  in  §  125  is  often  useful  in 
obtaining  the  first  derivative  of  airy  such  function. 

125.  If  u=f(x,  y),  du  =  ~dy  +  ^dx.  (1) 

dy  dx 

If  u  is  constant,  du=  0,  and  from  (1)  ive  obtain 

du 
cly_  _dx 
dx  ~      du 

dy 

in  which  —  and  —  are  partial  derivatives. 
dx  dy 

When  u  is  constant,  du  =  dxii  -f-  dyu  =  0  ;  but,  in  general, 
neither  dxu  nor  dyu  is  zero. 


1 .    y3  —  2  xry  -f-  bx  =  0  ;  find 
du 


Examples. 
dy 


Here 


dx 

du 


=  — 4ya;  +  &,      ^f  =  3y2-2x2; 
dx  dy 


dy 
dx 


du 

dx       Ayx  —  b 


du      3y2  —  2x2 
dy 


dy  _x2  - 

-  ay 

dx     ax 

-f 

dy 
dx 

-en 

Ka) 

dy  _ 

.y 

X 

fx  log  y  - 

-y\ 

dx 

\y  log  x  - 

-x) 

IMPLICIT   FUNCTIONS.  131 

2.  x3  +  yz  —  3  «.Ty  =  c  =  u. 

3.  £  +  £-1. 

4.  a;  log?/  —  ?/ logic  =  0. 

5.  /(aa; +  &?/)  =  c. 

Here        /(cw  +  by)  =  c  =  w  ; 

.-.  -^  =f'(ax  +  by)  a,     ^-A  =f'(ax  +  by)b; 

dy  _  _  a 
"da;         6 

6.  af-y-=0.  dy^y2-xy\ogy 

dx     x2  —  xy  log  a; 

7.  rf  +  8«f  — *  dy  =  _^  +  ay, 

da;         ?/2  -|-  ax 

126.  General  formulas  for  the  successive  derivatives  of  an 
implicit  function  can  be  easily  deduced  ;  but  they  are  so  com- 
plicated that,  in  practice,  it  is  generally  more  convenient  to 
differentiate  directly  the  first  derivative  to  obtain  the  second, 
aud  so  on,  than  to  use  these  formulas. 

Examples. 

1 .    Find  ^  when  y2  -  2  xy  +  c  =  0. 
dar 

Here        ^  =  _£_.  (1) 

dx     y  —  x 

Differentiating  (1),  we  have 

(y  —  x)^  —  y(~—  1 )     y  —  x-^ 
dry_ dx        \dx       J dx^ 

dx2  0/-a)2  (y-z)2 


132  FUNCTIONS    OF   TWO   Oli   MOBE   VARIABLES. 

From  (1)  and  (2), 

dry  _  y(y-2x) 

dx2      (y  —  n)3 

2 .    Given  y3  +  x3  -  3  axy  =  0  ;  show  that  ^  = 2_^/_. 

dxr  (y2  —  ax)3 

cl~v 
8.    Given  y1  —  2  aay  +  xr  —  c  =  0,  to  find  — \ 

dxr 

dry  _  (a2  —  1)  (y2  —  2  accy  ■+•  x2)  _  c(a2  —  1) 
da*2  (y  —  ax)3  (y  —  ax)3 


127.    Successive  Partial  Differentials   and  Derivatives.     If 

u=f(x,y),  x  and  y  being  independent,  dxii  and  dyu  are,  in 
general,  functions  of  both  x  and  y.  Differentiating  dxu  and  c^w 
with  respect  to  either  variable,  we  obtain  a  class  of  second  par- 
tial differentials.  Ity  differentiating  these  second  partial  differ- 
entials, we  obtain  a  class  of  third  partial  differentials  ;  and  so 
on.  In  finding  these  successive  partial  differentials,  we  regard 
dx  and  dy  as  constant,  since  we  may  suppose  x  and  y  to  change 
uniformly. 

The  successive  partial  differentials  are  represented  as  follows  : 


U^dx 

,dx 


d„  [  —  dx 
jlx 


dJ^dy 

w 

d3u 


d  fdu  ,\7' 
=  —   —  dx)  dx 
dx  \dx     J 

=  ±(^dxXhj 
dy  \dx     J 


=  — ,  dxr  ; 
dx- 


dx\dy-      J 


dru 
dx  dy 

d3u 


dyhlx 


dx  dy ; 

dy2dx ; 


d  f&hi 

dy\dy2 


dy2)dy 


dhi  ,  „ 
=  T^%3;  etc. 
dy3 


Hence -dxdy2  is  a  symbol  for  the  result  obtained  by  dif- 

dxdy- 

ferentiating  u  =/(.%*,  y)  three  times  in  succession :    first,  once 
with  respect  to  x,  and  then  twice  with  respect  to  y. 


PARTIAL   DIFFERENTIALS   AND   DERIVATIVES.       133 

The  symbols  for  the  partial  derivatives  are 
dhc      cPu      dhc    d?u       dhc 


da8    dxdy    dy2    dx*    da?dy 


etc. 


ir»o    t.p  ^/       \         dhc         dhc  dhc  dhc 

128.   If    u=f(x,y),     = ,     -  =  — ■ — ,    etc.; 

dydx      dxdy      dydar      dxrdy 

that  is,  if  u  be  differentiated  m  times  with  respect  to  x,  and  n 
times  with  respect  to  y,  the  result  is  the  same,  whatever  be  the 
order  of  the  differentiations. 

~f(x  +  bx,y)—f(x,y)' 
A# 


For  limit 


dx       Ax  =  0 

Regarding  this  expression  as  a  function  of  y,  finding  its  incre- 
ment, dividing  b}r  Ay,  and  remembering  that  the  difference 
between  the  limits  of  two  variables  is  equal  to  the  limit  of  their 
difference,  we  have 

d  (did 
dy\dxj 

[   limit  \~f(x+Ax>  y+M)—fix> .'/+ A-V )-/I*+A*.  >/)+f(x,  y)~ 
limit   \  Ax  =  0 


Ay  =  0  I  AlJ 

In  like  manner,  we  should  obtain  for  the  value  of  —  ( —  ]  the 


dx  \dy 

same  expression,  except  that  the  order  of  the  limits  would  be 
reversed  ;  but,  from  the  nature  of  the  process  of  passing  to  a 
limit,  it  is  evident  that  this  change  of  order  does  not  affect  the 
value  of  the  expression. 

TT  d  fdu\       d  fdxi\  dhi         dht  ,,x 

Hence      —    —    =  —    — )■>  or  = (1) 

dy\dxj      dx\dyj         dxdy     dydx 

Again,  since 

dhc         dhc 


dxdy      dydx 
dhc  cfxc 


dxdydx     dydx2 


(2) 


134  FUNCTIONS   OF   TWO    OK   MORE   VARIABLES. 

But,  from  the  principle  in  ( 1 ) ,  we  have 

d     d  fdu\  _  d     d  fdu\  dhi  dhi       ,„\ 

dx  dy\dxj      dy  dx^lxj^       dxdydx     dxhly 

From  (2)  and  (3), 

dhi      _    dhi 
dy  dx2     dx^dy 

This  method  of  reasoning  evidently  applying  to  all  cases,  the 
theorem  is  established. 


1 .  u  =  cos  (x  +  y)  ;  verify 

2.  u  =  x^y-  +  ay5 ;  verify 


Examples. 

d2u         dhi 


dydx     dxdy 
dhi         dhi 


dyhlx      dxdy2 

3.    u  =  log  (x  -f  y)  ;  find  the  first,  second,  and  third  partial 
derivatives. 


4.    u  =  tan    -;   verify 


dhi  dhc 


5.    u  =  sin  (bx5  -\-  ay5)  ;    verify 


dxhly     dy  dx2 

dhi,         dht 


dyhlx      dx  dy2 


129.    To  find  the  successive  differentials  of  a  function  of  two 
independent  variables. 

Let  u  =/(#,  y)  ;  then 

du  =  —  dx  +  —  dy.  (1) 

dx  dy   u  v  ' 

flit  (lit 

Differentiating  (1),  remembering  that  —  and  —  are,  in  gen- 
da;  dy 

eral,  functions  of  both  x  and  y  ;  and  that,  as  x  and  y  are  inde- 
pendent, dx  and  dy  mscy  be  regarded  as  constant,  we  have 

*-*(s*)+*(s*MS*MS* 


CHANGE   OF   THE   INDEPENDENT    VARIABLE.         135 

d?u  ,  o  .     dru    ,    7  dru    7    7     ,  dru  ,  «, 

=  T^  c7a^  +  -r^-  d*  c72/  +  t- 7-  chJ  dx  + 1-0  dV~ 
dx~  dxdy  dydx  dy~ 

«  *? rfar  +  2  -^-  dxdy  +  — ' dy8.  (2) 

dxr  dxdy  dy 

Differentiating  (2),  remembering  that,  in  general,  each  term 
is  a  function  of  both  x  and  y ;  and  that  the  total  differential  of 
each  is  the  sum  of  its  partial  differentials,  we  obtain 

d\c  =  tfida?  +  3  -^dx*dy+  3  J&-  dxdy2-\-—dy\ 
dx3  dx!dy  dxdy  dy3 

Differentiating  this  equation,  we  obtain  an  analogous  expres- 
sion for  dhc,  and  so  on. 

By  observing  the  analogy  between  the  values  of  dru  and  dsu, 
and  the  development  of  the  second  and  third  powers  of  a  bino- 
mial, the  formula  for  dnu  may  be  easily  written  out. 

130.  Change  of  the  Independent  Variable.   The  forms  of  the 

successive  derivatives  of  — "  used  thus  far  have  been  obtained 
dx 

upon  the  hypothesis  that  x  is  the  independent  variable,  and  that 
dx  is  constant.  In  applications  of  the  Differential  Calculus,  it 
is  often  desirable  to  change  the  independent  variable,  and  to 
regard  the  original  function,  or  some  other  variable,  as  the  inde- 
pendent variable.  TVre  proceed  to  find  the  forms  of  the  succes- 
sive derivatives  of  -^  upon  other  hypotheses  than  that  dx  is 
constant. 

131.  To  find  the  successive  derivatives  of  — ^  ;  that  is,  the  forms 

dx 

of  —  f  — 2.  ],  —  •- — [— 3- ),  etc.,  when  neither  x  nor  y  is  indepen- 
dx\dx/   dx  dx\dx/ 

dent. 

If  neither  x  nor  y  is  independent,  -J-  is  a  fraction  with  a  vari- 

dx 

able  numerator  and  denominator ;  and  we  have 
d  fdy\      dxdry  —  dydrx 


dx  \dxj  da? 


(1) 


13G  CHANGE   OF   THE   INDEPENDENT    VARIABLE. 

Differentiating  (1),  we  obtain 

d  _  d_  fdy\  _  _c^  fdxd2y  —  dy  d-x 


dx  dx\dxj      dx\  dx3 

_  (d5y dx — cfx dy) dx—3( d2y dx — drx dy) drx  ,,. 

dx5  (L) 

In  like  manner,  we  obtain  the  other  successive  derivatives. 

Cor.  1.    If  y  is  independent,  that  is,  if  dy  is  constant, 

d-y  =  0,  and  dzy  =  0, 

and  (1)  and  (2)  become 


—  (§y\  —  _  d~xcty 


dx\dxj  dx* 


(3) 


and  A. .  A  (clll)  _.  3(d2x)2dy-d3xdydx  ,^ 

dx  dx\clxj  dx'5 

Cor.  2.  If  dx  is  constant,  d2x  =  dsx=0,  and  (1)  and  (2) 
become 

jLfdlt)  =  clM.,  and  —  .  A  MA  =  ^M. 
dx\dxj      dx2"*  dx  dx\dxj      do,*3' 

which  agrees  with  §  74. 

Rem.  Hence,  to  transform  a  differential  expression  in  which 
x  is  'independent,  into  its  equivalent  in  which  neither  x  nor  y  is 

independent,  we  replace  — %  — ■«,  etc.,  bv  their  equivalents  upon 
dx-    dxr 

the  new  hypothesis,  which  are  found  in  (1),  (2),  etc. 

If,  in  the  transformed  expression,  a  new  variable  0,  of  which 
#is  a  function,  is  to  be  the  independent  variable,  in  the  general 
result  obtained  above  we  replace  x,  dx,  drx,  etc. ,  by  their  values 
in  terms  of  6  and  its  differentials. 

If  y  is  to  be  the  independent  variable,  in  the  given  expression 

d~u  d  v 
we  replace  ~,  — ^,  etc.,  by  their  equivalents  in  (3),  (4),  etc.;  or, 

in  the  general  result  first  obtained  above,  we  put  d2y  =  0,  d?,y  =  0, 
etc. 


EXAMPLES.  137 


Examples. 

1.  Given  ydry  +  dy2  +  dx2  =  0,  in  which  x  is  independent,  to 
find  the  transformed  equation  in  which  neither  x  nor  y  is  inde- 
pendent ;  also  the  one  in  which  y  is  independent. 

Dividing  both  members  by  dx~,  substituting  for  — *  the  sec- 
tor 
ond  member  of  (1),  §  131,  and  multiplying  both  members  by 

dx3,  we  have 

y(d2ydx  —  d2xdy)  +  dy-dx+  dx"J  =  0, 
in  which  neither  x  nor  y  is  independent. 

Putting  d-y  =  0,  and  dividing  by  —  dy3,  we  have 

cPx     dx3      dx      n 

y — o , =  0> 

dy      dy3      dy 

in  which  the  position  of  dy  indicates  that  y  is  independent. 

2.  Given  -2 :,  —  H — ,  =0,  in  which  x  is  indepen- 

dx-      1  —  :c-  dx      1  —  ar- 
dent, to  find  the  transformed  equation  when  x  =  cos6,  and  6  is 
independent. 

Substituting  for  — -;  the  second  member  of  (1),  we  have 

drydx  —  drxdy  _      x      dy  y      _  ~  ,   . 

dor3  1  —  or  cLk      1  —  ar 

Since       x  =  cos  0,  1  —  x2  =  s'urO, 

dx  =  —  sin  $  dd,  and  drx  =  —  cos  $  d62. 

Substituting  these  values  in  (1),  and  simplifying,  we  have 

dO- 
in  which  6  is  independent. 


138  CHANGE    OF   THE   INDEPENDENT    VARIABLE. 

df 


3.    Given  E 


1+^ 


in  which  x  is  independent,  to  find 


d'-y 

dx2 
the  value  of  R  when  x  =  pcos6,   y  =  ps'md,    and   0  is   inde- 
pendent. 

From  (1),  §  131, 

Baa    (<**»  +  <¥)*  (1) 

c?«  d-y  —  c/y  rf2rc 

in  which  neither  x  nor  ?/  is  independent. 
From  y  =  p  sin  0,  and  a;  =  p  cos#,  we  obtain 
dy  =  sin  0  dp  +y>  cos  0  c?0, 
rix  =  cos  0  dp  —  p  sin  0  d0, 
d2y  =  sin  0d2p  +  2  cos  Odd  dp  —  p  sin  0d02, 
and  d2x*  =  cos  0  d2p  —  2  sin  0  d6  dp  —  p  cos  0  d02. 

Substituting  these  values  in  (1),  and  simplifying,  we  have 


2  .  adp2        d2p 


4.  Given  —,-\ -+2/  =  0,  to  find  the  transformed  equa- 

dx2     x  dx 

tion  when  x2  =  4  z,  and  2  is  independent. 

Ans.     /ll!  +  fl!L  +  y=0. 

dz-      dz 

5.  Given  (l-x2)^- -a^  =  0,    to    find    the    transformed 

dx2        dx 

equation  when  x=  cos  z,  and  z  is  independent. 

Ans.  fl-O. 

(feT 

6.  Given  Z  =  XC^~^  X,  to  find  the  transformed  equation 

?/  ety  +  a;  dx 
when  a;  =  p  cos  0,  y  =  p  sin  0,  and  p  is  independent. 

dp 


CHAPTER   X. 
TANGENTS,  NORMALS,  AND  ASYMPTOTES. 

132.  The  Rectangular  Equation  of  the  Tangent  to  any  plane 
curve  at  (as',  y')  is 

dv'* 
ax 

For  line  (a)  passes  through  (x',  y')  ;  and,  by  §  16,  it  has  the 
slope  of  the  curve  at  (V,  y'). 

Cor.    "When  the  axes  are  oblique,  —  is  evidently  the  ratio 

of  the  sines  of  the  angles  which  the  curve  at  (x',  y')  makes  with 
the  axes  ;  hence,  in  this  case  also,  (a)  is  the  equation  of  the 
tangent. 

133.  The  Rectangular  Equation  of  the  Normal  to  any  plane 
curve  at  (a/,  y')  is 

y-y'=-(^(x-x').  (b) 

dy' 

For,  the  axes  beiug  rectangular,  line  (&)  is  perpendicular  to 
line  (a)  of  §  132,  and  therefore  to  the  curve,  at  (a;',  y'). 

Examples. 

1.    Find   the   equations  of  the    tangent   and   normal    to   the 
parabola  y-  =  2px. 

Here         dy  — P'    •  dy'  — P 

ax     y        ax'     y' 

-f-  represents  the  value  of  -f-  at  the  point  (x',j'). 
dx'  dx 


140  TANGENTS,    NORMALS,   AND   ASYMPTOTES. 

dv' 
This  value  of  —  substituted  iu   (a)  of  §  132,  and  (6)  of 
dx 

§  133,  gives 

2/-2/'=^(*-*%  (1) 

y' 

and  y  —  y'=  —  —  (x-x'), 

as  the  equations  of  the  tangent  and  normal  respectively. 
Since  y'2  =  2px',  equation  (1)  by  reduction  becomes 
yy'z=p(x  +  x'). 

2.  Find  the  equations  of  the  tangent  and  normal  to  the  circle 
x2  -\-  y2  =  r2. 

v' 

Ans.  yy'-\-xxl=  r2 ;  y —  y'=  —  (x  —  x'). 

3.  Find  the   equations  of   the   tangent  and  normal  to  the 
ellipse  ary2  -f  b2x*  =  a2b2. 

Ans.  a2yy'+  b2xx'=  a2b2 ;  y  —  y'=  -f-  (x  —  x') . 

b'-x 

4.  Find  the   equations  of  the  tangent  and   normal   to   the 
Iryperbola  a2y2  —  52ar  =  —  orb2. 

Ans.  a2yy' —  b2xx' =  —  orb2 ;  y  —  y'= f-(x  —  x'). 

blx' 

5.  Find  the  equations  of  the  tangent  and  normal  to  the  cis- 

soicl  y2— 

2a  —  x  A  ,       ,  ari(3a  — ar),  ,x 

Ans.  y  —  y '  =  ±  — * '-  (x  —  x')  ; 

J      J  (2a -xy   K  }  ' 

,  (2a-x')i   ,  ,, 

rs(3a  —  x') 

6.  Find  the  equation  of  the  tangent  to  y2  =  2x2  —  Xs  at  x  =  1. 

Ans.  y  =  %x  +  %;  y*=  —  \x  —  \. 

7.  Find  the  equation  of  the  normal  to  y2  =  6  x  —  5  at  y  =  5. 

Ans.  y  =  —  %%  +  ^-. 


LENGTH  OF  SUBTANGENT  AND  SUBNORMAL. 


141 


8.  Find  the  equation  of  the  tangent  to  the  hyperbola  referred 

to  its  asymptotes,  xy  —  m.  .  ?/ 

J  Ans.  y  =  --Lx+2y'. 

9.  Find  the  equation  of  the  tangent  to  the  cycloid 


y  l l'2v ?/' 

=  r  vers"1—  —  V  2  ry  -  y1.      Ans.  y  —  y'=^ ^-  {x  -  x') 


y 


134.  Length  of  Subtangent,  Subnormal,  Tangent,  and  Normal. 
Let   pt   be  the    tangent  at  the   point  p 
(sc\  y'),  and  rs  the  normal.     Draw  the 
ordinate  ra ;  then  tm  is  called  the  sub- 
tungent,  and  ms  the  subnormal. 

,dx' 


MP 

tm  = =  y 

tan  mtp      '    dy' 


Fig.  34. 


M  S 


.".subtangent  =  y 


,dx' 


ms  =  mp  tan  mps  =  y'  tan  mtp  =  y 


dx' 


.'.subnormal  =  y 


-„>dtf' 


dx' 


(1) 


(2) 


PT 


MP"  +  TM 


.'.tangent 


PS 


. ' .  normal 


(3) 


(4) 


Rem.  If  the  subtangent  be  reckoned  from  the  point  t 
(Fig.  34),  and  the  subnormal  from  m,  each  will  be  positive  or 
negative  according  as  it  extends  to  the  right  or  left. 


142 


TANGENTS,    NORMALS,    AND   ASYMPTOTES. 


Examples. 

1.    Find  the  values  of  the  subtangent  and  subnormal  of  the 
ellipse  cry2  +  b2xr  =  crb2. 

EM*.  ^*(-*-^l 
cly'  b~x'  x' 


Subn, 


J  dx' 


b2x' 
a2  ' 


2.    Find  the  values  of  the  subtangent  and  subnormal  of  tho 
parabola,  circle,  hyperbola,  and  cissoid. 

subn.  =p. 


Ans.  Parabola 
Circle : 


subt.  =  2x' ; 
subt.  =  —  K 


./■' 


H}Tperbola :  subt 
Cissoid : 


./•' 


subn.  =  —  x'. 
subn. 


or 


subt.  =^^^1;  subn.  ^^^ 
3  a  -  x' 


(■2a-x'y 

3.  Find  the  value  of  the  subtangent  of  the  logarithmic  curve 
y  =  a" ;  also  of  y2  =  3  x2  —  1 2  at  x  =  4.  .4?;s.  m  ■  3. 

4.  Find  the  values  of  the  subnormal  and  normal  of  the  cvcloid. 


Subnormal  =  ^/(2r  —  y)y  —  Vhb  -hd  =  ph  =  ed. 
Nonnal  =  pd  =  v  ed"  +  ep  =  V2  ry. 

Thus  the  normal  passes  through  the  foot  of  the  vertical  diam- 
eter of  the  generating  circle,  when  it  is  in  position  for  the 

point  to  which  the   normal 
t  s<z^~<L  is  drawn.     Moreover,  since 

dpb   is    a  right    augle,   the 
tangent  passes  through  the 
other  extremity  of  the  ver- 
^   tical  diameter.     This  prop- 
Fis- 35-  erty     furnishes     a     simple 

method  of  drawing  a  tangent  and  normal  to  the  cycloid  at  any 


POLAR   CURVES. 


143 


point.  Thus,  to  draw  a,  tangent  and  normal  at  p,  put  the  gen- 
erating circle  in  position  for  this  point,  and  draw  the  vertical 
diameter  bd.  The  lines  drawn  from  b  and  d  through  r  will 
be  respectively  the  required  tangent  and  normal. 

That  pb  is  tangent  to  the  cycloid  at  p  is  further  evident ; 
since,  when  the  generating  point  reaches  the  position  p,  it  is 
rotating  about  the  point  d,  and  is  therefore  moving  in  a  direc- 
tion perpendicular  to  dp. 

135.  Fundamental  Principle  in  the  Method  of  Limits.  Let 
a,  a,,  /?  and  ft,  be  any  four  variables,  so  related  that 

limit —  =  1 ,  limit  ~  =  1 ,  and  limit  „  =  c  : 
«i  Pi  P 

then,  since  «  =^  .^.A  =  ^I._^.fr, 
£      j8«,    ft      ft   a,    £' 

limit-  =  limit  -^  x  limit—  x  limit  &  =  limit -^i- 
0  ft  ax  0  ft 

Hence,  in  any  problem  concerning  the  limit  of  the  ratio  of 
two  variables,  either  may  be  replaced  by  any  other  variable,  the 
limit  ofichose  ratio  to  it  is  unity. 


136.  Length  of  Subtangent,  Subnormal,  Tangent,  and  Normal 
in  Polar  Curves.  Let  ox  be 
the  polar  axis,  and  p  any  point 
on  the  curve  mn.  Let  arc  pd 
=  As,  and  ob  =  1  ;  then  arc 
be  =  A0,  arc  pm  =  pA0,  and 
md  =  Ap.  Draw  the  chords  pm 
and  pd,  the  tangent  fz,  and 
rn  perpendicular,  and  zh  par- 
allel, to  op,  thus  forming  the 
right  triangle  pzh  ;  then 

PA0 


limit 
A9  =  0 


chord  pm 


=  1,    §48. 


Fig.  36. 


144 


TANGENTS,   NORMALS,   AND   ASYMPTOTES. 


and 


and 


limit 
As  =  0 


As 


chord  I'D 


1. 


imit      ; 
0=0    ' 


limit 
A 


limit 
A0  =  O 


pA6~ 
As 

"V 

As 


_  limit 
~  A6  =  Q 

limit 
A0  =  O 


chord  pm" 
chord  pd 


§  135. 


chord  pd 


The  limit  of  angle  mpd  is  evidently  hpz  ;  and,  in  the  isosceles 
triangle  pom,  the  limit  of  angle  pom  being  zero,  the  limit  of 
pmo  or  its  supplement  pmd  is  a  right  angle.  Hence  the  angles 
of  the  triangle  hpz  are  the  limits  of  the  angles  of  the  triangle 
mpd.  Therefore  the  ratios  of  the  sides  of  the  triangle  hpz 
equal  the  limits  of  the  ratios  of  pA6,  As,  and  Ap.  Hence  these 
sides  may  be  taken  as  pd6,  ds,  and  dp. 

Draw  ot  perpendicular  to  op,  and  produce  it  until  it  meets 
the  tangent  in  t.  Draw  also  the  normal  pa,  and  the  perpen- 
dicular on  upon  the  tangent.  The  lengths  pt  and  pa  are  called 
respectively  the  polar  tangent  and  polar  normal;  oa  is  the  polar 
subnormal,  and  ot  the  polar  subtangent. 

hp  _  p  d6 
hz       dp 

hp      p  dO 


tan  opt  =  tan  hzp 


sin  opt  =  sm  hzp  s=  — •  = 


ds 


(1) 
(2) 


From  triangle  hpz, 

ds2  =  dP2  +  p2d62. 

2d0 
ot  =  op  tan  opt  =  p* —  ; 
dp 


polar  subtangent  =  p 


,d$m 
dp 


oa  =  op  tan  opa  =  op  cot  opt  =  — 2 ; 

dd 

'.  polar  subnormal  =  — £• 
F  dO 


(3) 


(4) 


(5) 


=  v 


OP   +OT 


2     i        4f^2  . 


POLAR    CURVES.  145 


I  dff~ 

.'.  polar  tangent  =  p-y  1  +  /a2—-  (6) 

\  dp 


V— -'  ,  — -  -  i  dp 


.'.  polar  normal  =  a  \p2  -\ — —  •  (7) 

\  cW" 

ocW  phW  /Qv 

2>  =  on  =  op  sin  opt  =  p- —  =  —    ^  —  (8) 

<&      Vp2c70-  +  dp2 

137.  That  the  sides  of  the  triangle  pzh  (Fig.  36)  may  be 
taken  as  ds,  dp,  and  p  d9  can  he  proved  also  as  follows : 

When  the  generatrix  of  the  curve  is  at  p,  the  radius  vector 
is  increasing  in  length  in  the  direction  of  pr,  and  the  extremity 
of  the  radius  vector  drawn  to  p  is  moving  in  the  direction  of  ph, 
at  the  rates  at  which  the  generatrix  is  then  moving  in  these 
directions.  If,  at  p,  the  motion  of  the  generatrix  became  uni- 
form along  the  tangent  pz,  it  is  evident  that  any  simultaneous 
increments  of  its  distances  from  p  and  lines  ph  and  or  may 
be  taken  as  ds,  dp,  and  the  differential  of  the  arc  traced  by  the 
extremity  of  the  radius  vector  to  p,  which  equals  pd$;  for, 
if  ob  =  1,  b  describes  the  measuring  arc  of  the  variable  angle  6, 
and  p  is  moving,  as  op  revolves,  p  times  as  fast  as  b.  Hence 
ph  =  p  dO,  and  uz  =  dp,  if  pz  =  ds. 

Examples. 

1.  Find  the  subtangent,  subnormal,  tangent,  normal,  andjo, 
or  the  length  of  the  perpendicular  from  the  pole  to  the  tangent, 
of  the  spiral  of  Archimedes  p  =  a$. 

subnormal  =  —  =  a : 
dd 

subtangent  =  p2—  =  £- ; 
dp       a 


146  TANGENTS,    NORMALS,    AND    ASYMPTOTES. 

tangent  =  P%|l+p^:  =  P^l+|; 
normal  =  Vp2  +  a2 ;     p  =  — p- 


Vp-  +  a2 

2.    Find  the  subtaugent,  subnormal,  tangent,  and  normal  of 
the  logarithmic  spiral  p  =  a0. 

Ans.    subt.  sss     p     ;  subn.  =  p  log  a  ; 

log  a 

tan.  =  p-y/1  +  -, —2 ;       nor.  =  P  V 1  +  (log  a)  s. 

\         (loga)J 

Since  tanoPT=£—  = ,  this  curve  makes  the  same  angle 

dp       log  a 

with  every  radius  vector,  and  therefore  is  called  the  equiangular 

spiral. 

If  a  =  e,  tan  opt  =  1,  opt  =  -,  subtangent  =  subnormal,  and 
tangent  =  normal. 


3.    Find  the  subtangent,  subnormal,  and  p  of  the  lemniscate 

of  BernouiUi  p-  =  crcos  2  0. 

—  os  —  a2 

Ans.   subt.  =  — — " —  ;       subn.  = sin  2  6  ; 

a-  sin  26  p 

Ps  P3 

P  =  —  =  —. ;• 

Vp4  +  «4shr20      a* 


Rectilinear  Asymptotes. 

138.  A  Rectilinear  Asymptote  is  a  straight  line  that  has  the 
limiting  position  of  the  tangent  to  an  infinite  branch  of  a  curve. 
If  a  curve  has  no  infinite  branch,  it  evidently  can  have  no 
asymptote. 

If  X  and  Y  represent  the  intercepts  of  a  tangent  on  the  axes 
of  x  and  y  respectively,  from  the  equation  of  the  tangent  to 
any  plane  curve. 


ASYMPTOTES.  147 

«_«'=s^!(aj-a?f), 

we  obtain    X=x'  —  y'-—,  (1) 

cly' 

and  Y=y'-x'c^-r  (2) 

ax 

Now  if,  as  the  point  of  contact  (x',  y')  moves  out  along  an 
infinite  branch,  the  value  of  X  or  Y  or  the  values  of  both 
approach  finite  limits,  it  is  evident  that  these  limits  will  be  the 
intercepts  on  the  axes  of  an  asymptote  to  that  branch  of  the 
curve. 

Examples. 

1 .    Examine  ys=  Gx2  +  x3  for  asymptotes. 
Solving  for  y,  we  have 

As  x  =  oo,*  y  =  oo  ;  and,  as  x  =  —  oo,  y  =  —  oo. 

Hence  the  curve  has  two  infinite  branches,  one  in  the  first 


angle  and  another  in  the  third 

if 
£x'+x'2~       1  i  i 


X=x'--^- — -  =  -_-£- =-2  as  «'  =  ±a>; 


a? 


Y=y' ! =  -r-. rt—  2  as  x'  =  ±  oo. 


*x  =  aasy=<x  is  read  "x  approaches  a  as  its  limit  as  y  approaches 
infinity,  or  increases  without  limit."    A  variable  cannot  approacli  infinity 

as  its  limit.   For  example,  if  y  =— ,  and  x  =  0,  y  does  not  approach  infinity 

x 
as  its  limit;  for,  when  x  is  infinitely  near  0,  y  is  infinitely  large;   but  it 
doubles  its  value  while  x  decreases  by  half  its  own  value.    Hence,  as  x  =  0, 
the  difference  between  oo  and  y  must  always  be  many  times  as  great  as  y, 
however  great  y  may  become. 


148  TANGENTS,    NORMALS,    AND   ASYMPTOTES. 

Therefore,  the  line  y  =  x  +  2,  whose  intercepts  on  the  axes  of 
x  and  y  are  respectively  —2  and  2,  is  an  asymptote  to  each 
branch. 

2.    Examine  the  conic  sections  for  asymptotes. 

Neither  the  circle  nor  the  ellipse  can  have  an  asymptote,  since 
neither  has  an  infinite  branch. 

The  parabola  has  two  infinite  branches,  one  in  the  first  angle 
and  another  in  the  fourth. 

Here         X=x' =  —  x'  =  —  co  as  x'=  co  ; 

p 

,    px'     y'  .  ,  . 

and  1  =  ?/' r  =  — r  =  co  as  x'  =  go  . 

J      y'      2 

Hence  the  parabola  has  no  asymptote. 

The  hyperbola  has  four  infinite  branches,  one  in  each  angle. 
In  this  curve 

X=aj'-^  =  —  =  ± 0  as  x'  =  ± co  ; 

6V       x' 

1  =y' ■ —  = =  ±  0  as  x'  or  y  =  ±  co. 

cry'  y' 

Hence  the  asymptote  to  each  branch  passes  through  the  ori- 
gin.    To  determine  the  direction  of  these  asymptotes,  we  have 

cly'      b2x'       .  b  .    ,    b  ,  .    , 

-r~.  =  — — ;  =  ± =  =± —  as  x'  =  ±  ce. 

ax'     cry' 


a\h-- 

\        x'2 


b2x' 
Since  —^  is  positive  for  anv  point  in  the  first  or  third  angle, 

,         al  "  b     . 

and  negative  for  any  point  in  the  second  or  fourth,  y  =  -x  is 

the  asymptote  to  the  branches  in  the  first  and  the  third  angle,  and 
y  — x  to  those  in  the  second  and  the  fourth.  These  asymp- 
totes are  evidently  the  produced  diagonals  of  the  rectangle  on 
the  axes. 

3.    Prove  that?/  =  —  x  is  an  asymptote  to  each  of  the  two 
infinite  branches  of  y%  —  a5  —  x3. 


ASYMPTOTES.  149 

4.    Show  that  y-  =  ax*  has  no  asymptotes. 

139.  Asymptotes  Determined  by  Inspection  or  Expansion. 
From  the  definition  of  an  asymptote,  it  follows  that  it  is  a  line 
which  an  infinite  branch  of  a  curve  approaches  indefinitely 
near,  but  never  reaches.  From  this  view  of  asymptotes,  we 
can  often  determine  their  equations  by  inspecting  the  equation 
of  the  curve,  or  expanding  one  of  its  members.  Thus,  in  the 
cissoid 

y2  = ,y=±ooasa;  =  2«. 

2a  —  x 

Whence,  as=  2  a  is  an  asymptote  to  the  two  infinite  branches  of 
the  curve  ;  for  they  approach  indefinitely  near,  but  never  reach, 
this  line. 

In  x  =  log,,?/,  or  y  =  ax,x  =  —  cc  as  y  =  0.  The  axis  of  x  is 
therefore  an  asymptote  to  the  infinite  branch  in  the  second 
angle. 

Again,  the  equation  xy  —  ay—bx=0  may  be  put  in  the 
form, 

bx  ay 

y  = or  x  =  — 2—, 

x  —  a  y  —  b 

from  which  we  know  that  x  =  a  and  y  =  b  each  is  an  asymptote 
to  two  infinite  branches. 

The  method  of  examining  a  curve  for  asymptotes  by  solving 
its  equation  for  y,  and  then  developing  the  second  member  in 
descending  powers  of  as,  by  Maclauriu's  formula  or  some  other 
means,  will  be  illustrated  by  a  few  examples.* 

*  The  following  is  a  brief  view  of  another  method  of  examining  a  curve 
for  asymptotes.  For  a  fuller  treatment,  see  "Williamson's  Differential 
Calculus,  page  240. 

Let  the  equation  of  a  right  line  be 

y  =  nx+v;  (1) 

that  of  a  curve  of  the  nth  degree, 

/(*,2/)  =0;  (2) 

that  obtained  by  substituting  ux  +  v  for  y  in  (2), 

*(*)  =  0.  (0) 


150  TANGENTS,    NORMALS,    AND   ASYMPTOTES. 

Examples. 
1.    Examine  x*  —  xy2  +  ay2  =  0  for  asymptotes. 

Here        y=±x( ),==±»(l ■ 

\x  —  a/  \       x, 

\       2x      8ar 

From  the  first  form  of  the  value  of  y,  x=  a  is  evidently  an 
asymptote  to  two  branches  of  the  curve  that  lie  to  the  right  of  it. 
From  the  last  form  we  see  that  two  branches  of  the  curve  ap- 
proach infinitely  near  each  of  the  lines  y  =  ±  x  ±  -,  as  x  =  ±  oo. 

The   curve  therefore  has  three  asymptotes,   each  of  which  is 
asymptotic  to  two  infinite  branches. 

Equation  (3)  is  evidently  of  the  nth  degree,  and  its  n  roots  are  the 
abscissas  of  the  n  real  or  imaginary  intersections  of  (1)  and  (2).  If  two 
roots  of  (3)  be  equal,  two  points  of  intersection  of  (1)  and  (2)  will  coincide, 
and,  in  general,  (1)  will  be  a  tangent  to  (2).  From  Algebra,  we  know  that, 
as  the  coefficients  of  xn  and  x"-1  approach  zero  as  a  limit,  two  roots  of  (3) 
increase  without  limit.  Hence,  if  fi  and  v  in  (1)  have  such  values  as  ren- 
der these  coefficients  0,  (1)  has  the  limiting  position  of  a  tangent  to  an 
infinite  branch,  or  is  an  asymptote. 

For  example,  let  the  curve  be 

yz  —  ax2  +  xs.  ( 1 ) 
Substituting  [jux  +  v  for  y,  and  arranging  the  terms,  we  have 

(/xs  -1)  Xs  +  (3  fv  -a)  x*  +  3  ^x  +  v%  =  0.  (2) 

Two  roots  of  (2)  become  oo,  when  yu3  — 1  =  0,  and  3fjfiv  —  a  =  0;  that  is, 
when  fi=l,  and  j/=i«.     Hence,  y  =  x+^a  is  an  asymptote  to  (1). 

From  the  theory  of  equations,  and  this  theory  of  asymptotes,  the  fol- 
lowing are  obvious  conclusions : 

(a)  Any  asymptote  or  tangent  to  a  curve  of  the  third  degree  intersects 
the  curve  in  one,  and  only  one,  point. 

(b)  Any  asymptote  or  tangent  to  a  curve  of  the  nth  degree  cannot  meet 
it  in  more  than  n  — 2  points,  exclusive  of  the  point  of  contact. 


ASYMPTOTES   TO   POLAR    CURVES. 


151 


2.    Examine  y2  =  x2  — for  asymptotes. 


=  ±x    1- 


1 


=  ±.T  1-4  + 
ar 


2a2  + 


Hence  y  =  ±  x  are  the  two  asymptotes. 
3.    Examine  y5  =  ax2  —  Xs  for  asymptotes. 


Ans.  y  =  —  x-\- 


4.    Examine  y  —  c  + 


5.    Examine  y2 


(z-6)2 


for  asymptotes. 


x  —  a 


Ans.  y  =  c,  and  a  =  &. 
for  asymptotes. 

Ans.  x  =  a,  and  ?/  =  ±  (cc  +  a) . 


140.  Asymptotes  to  Polar  Curves.  If,  as  0  =  0',  p=oo,  and 
the  subtangent  od  =  os,  it  is  evident  that  sn, 
which  is  parallel  to  om,  is  an  asymptote  to  the 
infinite  branch  pk.  Hence,  to  examine  a  polar 
curve  for  asymptotes,  we  find  from  its  equa- 
tion the  values  of  6  which  make  p  =  ±co.  If 
the  corresponding  value  of  the  subtangent  is 
finite,  the  line  parallel  to  the  infinite  radius 
vector,  and  passing  through  the  extremity  of 
the  limiting  subtangent,  is  an  asymptote. 

Examples.  Fig.  37. 

1.   Examine  the  hyperbolic  spiral  p$  =  a  for  asymptotes. 

Here  p  —  -  ;   hence,  when  6  =  0,  p  =  oo,  and  subtangent  =  —a. 
9 

The  curve  therefore  has  an  asymptote  parallel  to  the  initial  line, 
and  at  the  distance  a  above  it. 


152  TANGENTS,    NORMALS,   AND   ASYMPTOTES. 

2.  Examine  p  cos  $  =  a  cos  2  0  for  asymptotes. 

Here  p  — ;   hence,  when  0  =  - ,  r  =  ±  go,  and  subtan- 

cos#  2 

gent  =  —  a. 

The  line  perpendicular  to  the  initial  line  at  the  distance  a  to 
the  left  of  the  pole  is  therefore  an  asymptote  to  two  infinite 
branches. 

3.  Examine  p2  cos  0  =  a2  sin  3  6  for  asymptotes. 

Ans.  The  perpendicular  to  the  initial  line  at  the  origin  is  an 
asymptote. 

4.  Show  that  the  initial  line  is  an  asymptote  to  the  lituus 
pV^=  a. 


Miscellaneous  Examples. 

1 .    At  what  angle  does  y2  =  1 0  x  intersect  x2  -f-  y2  =  144  ? 

Ans.   71°0'58". 


2.   Find  the  subnormal  of  the  curve  y2=  2 a2 logic. 


Ans.  — 


3.   Find  the  equation  of  the  tangent  to  the  curve 

a?{x  -\-y)=  a2(x  —  y)  at  the  origin. 

Ans.  y  =  x 


4.  Find  at  what  angle  the  curve  y2—2ax  cuts  the  curve 

a?  —  3 axy  -+-ys  =  0. 

Ans.  cot_1V4. 

5.  Find  the  normal,  subnormal,  tangent,  and  subtangent  of 

X  X 

the  catenary  y  =  -  (eu  +  e  ") . 

2x  _2x  2 

Ans.  Subn.=-(ea—  e    a)  ;     norm.  =  iL- 
4 v  '  a 

subt.  = —  ;  tan.= — ~ 

■>Jy2  —  a2  -\/y2—a2 


6.    Find  the  sub  tangent  of  the  curve  y2  = 


EXAMPLES.  153 

a? 

a  —  x 

A        2x(a  —  x) 

Ans.  * '— 

3  a  —  2  x 


7.  Examine  y*{x  —  2a)  =  xs  —  a8  for  asymptotes. 

Ans.  x  =  2a\  y  =  ±  (x  +  a) . 

8.  Examine  y(a2  —  a;2)  =  b~(2x  +  <?)  for  asymptotes. 

Ans.  y  =  0  ;  #  +  ft  =  0  ;  x  =  a. 

9.  Examine  for  asymptotes  the  folium  of  Descartes, 

xs  +  y3-haxy  =  0. 

Ans.  y  =  —  x  —  a. 

10.  Examine  (y2  —  l)y—  (x2  —  A)x  for  asymptotes. 

Ans.  y  =  x. 

. .      t.        •        o      a2(«-a)(x-3o)   -  ,   , 

1 1 .  Examine  y  =  — * — ^ -  for  asymptotes. 

ar  —  2  ckb 

-4us.  #  =  2a;  #  =  0;  y  =  ±a. 

12.  Examine  the  hyperbola  p  =  — i — — — —  for  asymptotes. 

e  cos  0—1 

13.  Find  the  length  of  the  perpendicular  from  the  pole  upon 
the  tangent  to  the  lituus  p V#  =  a.  a  _        2a- p 

Vp4  +  4a4 

14.  In  the  hypocycloid  xi  +  yi  =  a%,  prove  that  the  portion 
of  the  tangent  intercepted  between  the  axes  equals  a. 

15.  Give  the  different  methods  of  drawing  a  tangent  to  any 
plane  curve  at  a  given  point. 


CHAPTER  XL 


DIRECTION  OF  CURVATURE,  SINGULAR  POINTS,  AND  CURVE 

TRACING. 

141.   Direction  of  Curvature.   A  curve  is  concave  upward  or 
downward  at  any  point,  according  as  in  the  immediate  vicinity 
of  that  point  it  lies  above  or  below  the  tangent 
at  that  point. 

When  a  curve,  as  o&,  is  concave  upward,  it 

•  dv 

is  evident  that  — ,  the  slope  of  the  curve,  in- 

dx  p 

creases  as  x  increases  ;  hence,  — 4(,  the  deriva- 
tive of  -^,  is  positive  (§  72). 
dx     7 
ay 


Fig.  38. 


AVhen  a  curve, 


is  negative. 


as  cd,  is  concave  downward,  —  decreases  as  x  increases,  and 

d2y 

dx2 

Hence,  the  curve  j'=f(x)  is  concave  xipivard  or  doivnward, 

at  any  point  (x,  y),  according  as  — ^  is  positive  or  negative. 

In  the  polar  system,  a  curve  is  said  to  be  concave  or  convex 
toward  the  pole  at  an}r  point  according  as  in  the  immediate 
vicinity  of  that  point  it  lies  on  the  same 
side  of  the  tangent  as  the  pole,  or  on  the 
opposite  side.  From  the  figure,  it  is 
evident  that,  when  the  curve  is  concave 
toward  the  pole,  p  or  od  increases  as  p 

increases,    and    —    is    positive    (§  31). 
dp 

When  the  curve  is  convex   toward   the 

pole,  p  decreases  as  p  increases,  and  -~ 

dp 
is  negative. 

Hence,  a  polar  curve  is  concave  or  convex  toivard  the  pole 

according  as  — L  is  positive  or  negative, 
dp 


EXAMPLES.  155 

When  the  equation  of  the  curve  is  given  in  terms  of  p  and  0, 
we  findp  in  terms  of  p  by  use  of  (8)  of  §  136. 


Examples. 
1 .    Find  the  direction  of  curvature  of  y  =  a?  +  2x  +  5. 

&y 

dx2 


d2v 
Here  — 4  =  2  ;  hence  the  curve  is  concave  upward. 


2.  Find  the  direction  of  curvature  of  y  —  a  +  c(x  -f  &)8. 

Here  — 4=  Gc(x  +  b)  ;  hence  the  curve  is  concave  upward  or 
dxr 

downward  at  (x,  y) ,  according  as  x  >  or  <  —  b. 

3.  Find  the  direction  of  curvature  of  y  =  x9  —  3  xr  —  9  x  -\-  0. 
.4ns.  Concave  upward  or  downward  according  as  a;  >  or  <  1 . 

4.  Find  the  direction  of  curvature  of  x  =  logay,  and  y  =  since. 


Here 


5.    Find  the  direction  of  curvature  of  the  lituus  p20  =  a2. 

dp  _         a2  pz  . 

d$~      2pb2~      2  a2' 

V4it4+p4' 


r   ^  da2 


db2 

■  dp  =  2a2(iai-pi) 
"dp        (4a*+p*)J  ' 

Hence  this  spiral  is  concave  or  convex  toward  the  pole  at  (9,p) 
according  as  p  <  or  >  a v2. 

6.    Find  the  direction  of  curvature  of  the  logarithmic  spiral 
p  =  a9. 

Here        p  — —       p        — ;     .'.-&  =  —  — • 

V 1  +  (logo)8  dP      Vl  +  (log  a)2 

Hence  the  curve  is  concave  toward  the  pole. 


156  SLNGULAE   POINTS. 

Singular  Points. 

142.  Singular  Points  of  a,  curve  are  those  which  have  some 
peculiar  property.  Such  points  are  :  first,  Points  of  Inflexion ; 
second,  Multiple  Points ;  third,  Conjugate  Points ;  fourth,  Stop 
Points. 

143.  Points  of  Inflexion.  A  point  of  inflexion  is  a  poiut  at 
which  a  curve  changes  its  direction  of  curvature.  Hence  a 
x  tangent  at  a  point  of  inflexion  intersects 

the  curve.     Thus  the  tangent  at  p,  a  point 
of  inflexion,  cuts  the  curve  at  p. 

At  a  point   of   inflexion   on   y=f(x), 

— ^  must  evidently  change  its  sign,  and 
dx~ 

therefore  pass  through  0  or  go.     Hence, 

—         d~v 
o  x    if  __£  be  found  in  terms  of  aj,  the  roots  of 

Fig.  40.  d^r 

-J.  =  0  or  go  are   the  critical   values   of   x   to   be    examined. 
dx2 

d2v 
If  —?.   changes  its  sign  as  x  passes  through  any  one  of  these 

dx2 
values,  this  value  is  the  abscissa  of  a  point  of  inflexion.* 

Examples. 

1.  Examine  a?  —  3  bx2  +  a?y  =  0  for  points  of  inflexion. 

The  root  of  — -0  =     *■    ~ — '-  =  0  is  b  ;  and  — ^ — ~*      evidently 
dxr  cv  cr 

changes  its  sign  as  x  passes  through  b;  hence  [6,  — —  j  is  a 
point  of  inflexion.  ^  ' 

2.  Prove  that  the  points  in  which  y  =  csin-  cuts  the  axis  of 
x  are  all  points  of  inflexion. 

*  On  one  side  of  a  point  of  inflexion,  the  slope  of  a  curve  is  increasing ; 
and,  on  the  other,  it  is  decreasing :  hence  a  point  of  inflexion  is  a  point  of 
maximum  or  minimum  slope,  and  the  method  of  finding  such  a  point  is 

seen  to  be  that  of  finding  a  maximum  or  minimum  of  ~^- 

dx 


MULTIPLE   POINTS.  157 

The  roots  of  — \  =  — _sin-=3  0  are  0,  air,  2 air,  3 an-,  etc.   As 
dor  cr       a 

x  passes  through  each  of  these  values,  —2  changes  its  sign;  hence 

dar 

0,  air,  2 air,  Sair,  etc.,  are  abscissas  of  points  of  inflexion. 

3.  Examine   the   witch   of   Agnesi,  x2y  =  4a2(2a  —  y),   for 

points  of  inflexion.  .         ,  ,   „       ,-    ■>  ,* 

1  Ans.   (±faV3i  la)- 

Xs 

4.  Examine  y  = =  for  points  of  inflexion. 

a-  +  x- 

Ans.   (0,  0),  (a  V3,  f«V3),  and  (-aV3,  -|aV3). 

144.  To  test  curves  given  by  their  polar  equations  for  points 

of  inflexion,  we  find  the  roots  of  —  =  0  or  oo.     If  —  changes 

dp  dp 

its  sign  as  p  passes  through  any  one  of  these  critical  values, 
this  value  is  the  radius  vector  of  a  point  of  inflexion.*     Thus, 

in  the  lituus,   (28°  38',  a  V2)  is  a  point  of  inflexion  ;    for  -f. 

dp 
dp 
changes  its  sign  as  p  passes  through  a  V2,  the  root  of  -j-  =  0. 

(See  §  141,  Ex.  5.) 

Multiple  Points. 

145.  A  Multiple  Point  is  one  through  which  two  or  more 
branches  of  a  curve 

pass,  or  at  which 
they  meet.  A  mul- 
tiple point  is  double  b-  d- 
when  there  are  only 
two  branches,  triple 
when  only  three,  and 
so  on. 

A  multiple  point  at  which  the  branches  intersect  (Fig.  a)  is 
called  a  Multiple  Point  of  Intersection. 

*  A  point  of  inflexion  on  a  polar  curve  evidently  corresponds  to  a 
maximum  or  minimum  of  p. 


Fig.  41. 


158  SINGULAR   POINTS. 

A  multiple  point  through  which  two  branches  pass,  and  at 
which  they  are  tangent  (Figs,  b,  c)  is  an  Osculating  Point. 

A  multiple  point  at  which  two  branches  terminate,  and  are 
tangent  (Figs,  d,  e)  is  a  Cusp.  A  cusp  or  osculating  point  is 
said  to  be  of  the  first  or  the  second  species,  according  as  the 
two  branches  are  on  opposite  sides  (Figs.  b,  d)  or  the  same  side 
(Figs,  c,  e)  of  their  common  tangent. 

A  Conjugate  Point  is  one  that  is  entirely  isolated  from  the 
rest  of  the  real  locus.  Hence,  in  an  algebraic  curve,  a  conju- 
gate point  is  a  multiple  point  formed  by  the  intersection  or 
meeting,  in  the  plane  of  the  axes,  of  imaginary  branches  ;  that 
is,  of  branches  lying  outside  of  the  plane  of  the  axes.  Since 
an  odd  number  of  roots  of  an  algebraic  equation  cannot  be 
imaginary,  an  even  number  of  imaginary  branches  must  inter- 
sect or  touch  in  a  conjugate  point  of  an  algebraic  curve. 

146.  From  the  definitions  given  above,  it  follows  that,  at  a 

multiple  point  of  intersection,  -&  must  have  two  or  more  un- 

dx 

equal  real  values  ;  that,  at  a  point  of  osculation  or  a  cusp  it 
must  have  two  equal  real  values ;  that,  at  a  conjugate  point  on 
an  algebraic  curve,  it  must  have  two  or  more  values  which  .are 
imaginary,  unless  the  tangents  to  the  imaginary  branches  at  the 
conjugate  point  lie  in  the  plane  of  the  axes. 

Hence  at  any  multiple  point  -2  has  two  or  more  values. 

dx 

147.  If  f(x,  y)  =  u  =  0  be  the  algebraic  equation  of  a  curve 
freed  frpm  radicals  and  fractions,  at  any  multiple  point  upon  the 
curve, 

du 

dy  __  _  das  _  0      ,  du  _du  _  ~ 
dx         du      0'       dx     dy 

dy 

For,  at  anv  multiple  point,  — ,  or  the  ratio  of  —  to  — ,  must 

dx  dx      dy 

have  two  or  more  values  (§  146).     But,  from  the  form  of  the 


MULTIPLE   POINTS.  159 

equation  of  the  curve,  neither  —  nor  —  can  contain  radicals  or 

dx         dy 

fractions  ;  hence  their  ratio  can  have  two  or  more  values  for  the 
same  values  of  x  and  y,  only  when  it  assumes  the  indeterminate 

form  — 
0 

148.  Examination  of  a  Curve  for  Multiple  Points.  To  examine 
a  curve  for  multiple  points,  put  its  equation  in  the  required  form 
f(x,  y)=:u  =  0,  and  find  the  sets  of  values  of  x  and  y  that  will 

satisfy  the  equations  —  =  0  and  — =  0.     Of  these  sets,  those 
dx  dy 

which  satisfy  the  equation  of  the  curve  give  the  points  to  be 
examined. 

Let  (#',  y')  be  one  of  these  points  ;  then 
du 
dtf  =  _  dd_  _  0 
dx'         du      o' 

dy' 

which  is  evaluated  according  to  the  method  of  §  83. 

djf 
dx' 
general  a  multiple  point  of  intersection. 

du' 

II.  If  -+-  has  two  equal  real  values,  (x1,  y')  is  in  general 

dx' 
either  an  osculating  point  or  a  cusp. 

du' 

III.  If  all  the  values  of  —  are  imaginary,  (x',  y')  is  a  con- 

dx 

jugate  point. 

The  following  considerations  enable  us  to  discover  more 
exactly  the  nature  of  these  points  : 

In  Case  I.,  if  the  values  of  y  or  -^  are  real  for  x  =  x'—h 

dx 

and  x  =  x'+h,  h  being  very  small,  (x',  y')  is  a  multiple  point  of 
intersection  ;  if  real  for  neither,  (a;',  y')  is  a  conjugate  point  at 
which  the  imaginary  branches  are  parallel  to  the  plane  of  the 
axes. 


I.    If  —  has  two  or  more  unequal  real  values,  (»',  y')  is  in 

r7<v' 


160  SINGULAR   POINTS. 

In  Case  II.,  if  the  values  of  y   or  -^  are  real  for  x  =  x'—h 

dx 

and  x  =  x'-\-h,  (#',  y')  is  a  point  of  osculation  ;  if  real  for  only 
one  of  these  values  of  x,  (a;',  y')  is  a  cusp  ;  if  real  for  neither, 
(#',  y')  is  a  conjugate  poiut. 

dy' 
In  some  curves,  and  especially  when  —  =  oo,  it  is  better  to 

inspect  the  values  of  x  or  —  for  y  =  y'  —  h  and  y  =  y'+  h. 

dy 

To  determine  the  species  of  a  cusp  or  point  of  osculation, 

find  — •-{,  and  (x',y')  will  be  of  the  first  or  the  second  species, 

cl'v' 
according  as  the  two  values  of  — »-  have  opposite  signs  or  the 

same  sign.  Or  we  may  compare  the  ordiuates  or  abscissas  of 
adjacent  corresponding  points  on  the  branches  and  on  their 
common  tangent.* 


*  The  following  is  a  brief  view  of  another  method  of  examining  a  curve 
for  multiple  points.  For  a  fuller  treatment,  see  Williamson's  Differential 
Calculus  or  Salmon's  Higher  Plane  Curves. 

Let  the  equation  of  a  curve  of  the  nth  degree  be 

f(x,H)=0;  (1) 

that  of  a  right  line  through  the  origin, 

y  =  ltx;  (2) 

that  obtained  by  substituting  fix  for  y  in  (1), 

<M-r)=0.  (3) 

If  (1)  contains  no  constant  term,  its  locus  evidently  passes  through  the 
origin.  If  (1)  contains  no  constant  term  nor  any  term  of  the  first  degree, 
two  roots  of  (3)  are  0.  Hence  two  points  of  intersection  of  (1)  and  (2)  are 
at  the  origin;  and  the  origin  is  a  double  point.  If,  in  addition,  (1)  con- 
tains no  term  of  the  second  degree,  three  roots  of  (3)  are  0,  and  the  origin 
is  a  triple  point. 

Hence,  when  the  origin  is  a  multiple  point,  this  fact  is  evident  from  the 
equation  of  the  curve.  To  examine  a  curve  for  multiple  points  not  at  the 
origin,  change  the  reference  of  the  locus  to  new  parallel  axes,  using  the 
formulas  x  =  m  +  xv  and  y  =  n  +  yv  If  m  and  n  can  be  so  determined  that 
the  resulting  equation  will  contain  no  constant  term  nor  any  term  of  the 
first  degree  in  x1  or  yv  (m,  n),  or  the  new  origin,  is  a  double  point.  If  m 
and  n  can  be  so  determined  that  the  new  equation  will  contain  no  constant 


EXAMPLES.  161 


Examples. 


1.    Examine  x*  +  cixhf  —  atf  =  0  for  multiple    points. 

Here         u  =  x*  +  a.ry  —  aif  =  0  ;  ( 1 ) 

.  • .  —  =  4  a?  +  2  aa?y,  and  —  =  ax2  —  3  a  y2 ; 

dy  _  4  .r'1  4-  2  ax//  . 

dx       Say2  — ax2 

Placing  the  partial  derivatives  equal  to  zero,  we  have 

»(2oj2  +  ay)  =  0,  (3) 

and  x2  —  oy2=0.  (4 ) 

Solving  (3)  and  (4),  we  obtain  the  following  three  sets  of 
values  for  x  and  y, 

x  =  0,  -|-aV3,  and  —  £aV3, 

y  =  0,  —  £a,  and  -{a. 

Only  the  first  set  of  values  will  satisfy  (1)  ;  and  (0,  0)  is  the 
only  point  to  be  examined.     From  (2)  we  have 

dy~]    _  4  x*  +  2  axy~ 
dx  ^q     3  ay2  — ax2 

Hence  the  origin  is  a  triple  point  at  which  the  inclinations  of 
the  branches  are  respectively  0,  -^7r,  and  f  tt. 

From  (1),  

x=  ±\  —  ^ay±%y  V±ay  +  a2.  (5) 

From  (5) ,  we  see  that  four  values  of  x  are  real  when y  =  —  h, 
and  two  when  y  =  +  h  ;  hence  each  of  the  three  branches  passes 

term  nor  any  of  the  first  or  second  degree  in  x1  or  yv  (m,  n)  is  a  triple 
point. 

From  this  method,  it  is  evident  that  a  curve  of  the  third  degree  can 
have  only  one  multiple  point;  one  of  the  fourth  degree,  only  two  double 
points  or  one  triple  point ;  one  of  the  fifth  degree,  only  two  double  points 
or  one  triple  and  one  double  point. 


=  0  and  ±  1.  §  83. 

0,0 


162 


SINGULAR   PODSTTS. 


through  the  origin,  which  is  therefore  a  triple  point  of  inter- 
section. The  general  form  of  the  curve  at  the  origin  is  shown 
in  Fig.  a  on  page  157. 

2.  Examine  if  =  x2  (a2— x2)  for  mul- 
tiple points. 

Ans.   (0,  0),  Fig.  42,  is   a   double 


Fig.  42. 


point  of  intersection  ;  — •- 
dx 


±1. 


0,0 


3.  Examine  Xs  —  3  axy  +  if  =  0  for  multiple  points. 

Ans.  (0,  0)  is  a  double  point  of  intersection  ;  -£ 

dx 

4.  Examine  y2  (a2  —  x2)  =  x4  for  multiple  points. 

dy~ 

dx 


=  0  and  oo. 


=  ±0, 


and  from  its  equation  we  see  that  the  curve  extends  through  the 
origin,  and  is  symmetrical  with  respect  to  the  axis  of  x;  hence 
the  origin  is  a  point  of  osculation  of  the  first  species. 

5.    Examine  y2  =  arx*  for  multiple  points. 
<*3f 


dx 


±0, 


and  from  its  equation  we  see  that  the  curve  consists  of  two 
branches  symmetrical  with  respect  to  the  axis  of  x,  and  extend- 
ing from  the  origin  to  the  right ;  hence  the  origin  is  a  cusp 
of  the  first  species,  the  common  tangent  being  the  axis  of  x 
(Fig.  43). 


=  ±oo,    and 


6.  Examine  y3  =  ax2  —  x*  for  multiple  points. 
Here    (0,  0)    is    the   only   critical   point.     — 

y  =  (ace2  —  ar)  s  shows  that  there  is  a  branch  on  each  side  of  the 
axis  of  y,  neither  of  which  extends  below  the  origin,  which  is 
therefore  a  cusp  of  the  first  species. 

7.  Examine  y'2  =  x  (x  +  a)2  for  multiple  points. 

dy 


Here    (  —  a,  0)   is    the    critical   point,    and  — 
hence  (  —  a,  0)  is  a  conjugate  point. 


CURVE   TRACING.  163 


8.    Examine  ahf  —  2  abafy  —  af  =  0  for  singular  points. 


Here  (0,  0)  is  the  critical  point,  and  '  ' 

dx 

When  a?  =  0  +  &, 


=  ±0. 

0,0 


bit1  ,      llPh*  +  ah5 
a-       \       a 
in  which  both  the  values  of  y  are  real,  one  greater  and  the  other 
less  than  0. 

When  x  =  0  —  h, 


blr  ,      lb-V  -  ah5 
a-       \        cr 


in  which  both  the  values  of  ?/  are  real  and  greater  than  0  when 
1)  is  small. 

Hence  the  origin  is  a  point  of  osculation,  and  a  point  of  in- 
flexion on  one  branch. 

149.  A  multiple  point  at  which  two  or  more  branches  termi- 
nate, and  have  different  tangents,  is  a  Shooting  Point.  A  Stop 
Point  is  a  point  at  which  a  single  branch  of  a  curve  terminates. 

From  the  law  of  imaginary  roots  of  algebraic  equations, 
neither  a  shooting  point  nor  a  stop  point  can  occur  on  an  alge- 
braic curve. 

150.  Curve  Tracing.  The  most  rudimentary  method  of  trac- 
ing a  curve  is  to  find  from  its  equation  such  a  number  of  its 
points  that,  when  located,  these  points  will  clearly  indicate  the 
form  of  the  curve. 

This  method  is  laborious  ;  and  our  present  object  is  to  utilize 
the  principles  heretofore  developed  to  determine  directly  from 
its  equation  the  general  form  of  a  curve,  especially  at  such  points 
as  present  any  peculiarity,  so  that  the  curve  may  be  traced  with- 
out the  labor  of  the  first  method. 

To  trace  a  curve  from  its  rectilinear  equation,  the  following 
general  directions  will  be  found  useful. 

Solve  its  equation  for  y  or  x,  and  determine  any  lines  or  points 
with  respect  to  which  the  curve  is  symmetrical. 


164 


CURVE   TRACING. 


Find  the  points  at  which  the  curve  cuts  the  axes,  and  deter- 
mine its  limits  and  infnfTte  branches. 

Determine  the  positions  of  its  asymptotes,  and  on  which  side 
of  each  the  infinite  branches  lie. 

Find  its  maxima  and  minima  ordiuates,  and  the  angles  fit 
which  it  cuts  the  axes. 

Determine  its  direction  of  curvature,  points  of  inflexion,  and 
multiple  points. 

Examples. 

1.    Trace  the  curve  whose  equation  is  y2  =  a2x3. 
Here  y  =  ±  a#i,  and  the  curve  is  s}7mmetrical  with  respect  to 
the  axis  of  x. 

When  x  =  0,  y  =  0,  and  the  curve  meets  the  axes  at  (0,  0). 
When  x  <  0,  y  is  imaginaiy  ;  but  when  x  >  0,  y  is  real.    Hence 
there  is  one  infinite  branch  in  the  first  angle, 
and  another  in  the  fourth. 


dy 
dx 


3<rx2 
~Ty~ 


oo  when  x  ==  oo 


hence  the  curve  has  no  asymptote. 
dy 
dx 


^^  =  -at  (0,0)  ; 
2y        0        V       ; 


Fig.  43. 


dy~~\      _   ,  3  ax* 


but  ^  I   =  ± 

dx  o,o 


2 


=  ±0. 


Hence  the  two  symmetrical  branches  terminating  at  the  origin 
are  tangent  to  the  axis  of  x  at  that  point,  and  the  origin  is  a 
cusp  of  the  first  species. 

Since  — \  =  ± ,  the  upper  branch  is  concave  upward,  and 

dx"  4  -yjx 


the  lower  one  concave  downward, 
shown  in  Fia;.  43. 


The  form  of  the  curve  is 


2.    Trace  the  curve  yi=2  oaf  —  x3. 
Here  y  =  (2  ax2  —  ar3)  s. 

For  y  =  0,  x  =  0  and  2  a  ;  hence  the  curve  cuts  the  axis  of  x 
at  the  origin,  and  2  a  at  the  right  of  it. 


EXAMPLES.  165 

For  each  real  value  of  x,  y  has  one,  and  only  one,  real  value, 
which  is  +  or  —  according  as  x<  or  >2a.  Hence  there  is 
one  infinite  branch  in  the  second  angle  and  another  in  the 
fourth. 

To  find  the  equation  of  the  asymptote,  we  have 
y  =  (2  aa?  -  a?)*  =  -  as  (l  -  —  Y 

A      2a      4a2  \  m 

— %1^ii"»"""7      (1) 

when  x  is  numericall}'  greater  than  2  a. 

Hence  the  equation  of  the  asymptote  to  each  infinite  branch  is 

y  =  -x  +  ±a.  (2) 

From  (1)  and  (2),  it  is  evident  that  the  infinite  branches  lie 
between  the  asymptote  and  the  axis  of  x. 

chi      iax  —  Sx2      at,  4  . 

-a  = =  0  when  a!  =  |a; 

dx  3  y- 

.-.  (2  ox-2  —  a?)*]|o,  or  £  a-v^i, 
is  a  maximum  ordinate. 


da; 


=   00 
2o,0 


hence  the  curve  is  perpendicular  to  the  axis  of  x  at  (2  a,  0), 

dii      4  ax  —  3  x2      0       ,  A 

-£  = =  - ,  when  x  =  w  =  0. 

dx  3/  0  J 

By  evaluating  we  find  that 


clx 


±  oo  ; 


hence,  as  the  curve  does  not  extend  below  the  axis  of  x  to  the 
left  of  x  =  2  a,  and  y  has  one,  and  but  one,  real  value  for  each 
value  of  x,  the  origin  is  a  cusp  of  the  first  species,  the  two 
branches  being  tangent  to  the  axis  of  y. 


166 


CUUVE   TRACING. 


dry 
(fee8 


9a$(2a  —  x)l 

which  is  +  or  — ,  according  as  x  >  or  <  2«  ;  hence  (2  a,  0)  is 

a  point  of  inflexion,  to  the  right 
of  which  the  curve  is  concave 
upward,  and  to  the  left  down- 
ward. The  form  of  the  curve 
is  given  in  Fig.  44. 


Fig.  44. 


3.    Trace  the  curve 

y2(x?  —  a2)  =  x\ 

Since  its  equation  involves 
only  even  powers  of  x  and  y, 
the  curve  is  symmetrical  with 
respect  to  each  axis.  Hence,  if 
we  determine  the  part  of  the  locus  that  is  in  the  first  angle, 
the  symmetry  of  the  curve  will  give  us  the  other  three  parts. 

Solving  for  y,  we  have 

x> 

y  —  ± • 

J  (x2-a2)h 

When  x  =  0,  y  =  0  ;  but,  for  other  values  of  x  between  —a 
and  -f-  a,  y  is  imaginary  ;  hence  (0,  0)  is  a  conjugate  point,  and 
the  locus  in  the  first  angle  lies  to  the  right  of  x=  a. 

As  x  =  a  from  a  value  greater  than  a,  y  =  ao  ;  hence  there  is 
one  infinite  branch  in  the  first  angle  to  which  x  =  a  is  the  asymp- 
tote. "When  x  =  go,  y  =  go  ;  and  there  is  a  second  infinite  branch 
in  the  first  angle. 

To  find  the  equation  of  the  other  asymptote,  we  have 


y  =  ±. 


x  1 


=  ±x   1- 


{*£+-} 


=  ±a;[l  +  £-  + 

2  x- 

when  x  is  numerically  greater  than  a  ;  hence  y  =  x  is  the  equa- 
tion of  the  other  asymptote. 


EXAMPLES. 


1G7 


Evidently  the  curve  lies  above  this  asymptote.      Hence  the 

branch  in  the  first  angle  lies  above  y  =  X,  and  to  the  right  of 
as  =  a. 


dy  _  2  xs  —  xy2  _  ^ 
dx      y(x2  —  a1) 

when   y2  —  2  a;2,    or   x  =  ±  aV2  ; 
hence   2  a  is  a  minimum  ordi- 
nate. 

In  the  first  angle, 

d2>/  =  ar(x2  +  2a2) 
dx2        (xr  —  a2)  5 


Fig.  45. 

The  form  of  the 


As  this  is  -f-  when  x  >  a,  this 
branch  in  the  first  angle  is  concave  upward, 
curve  is  given  in  Fig.  45. 

X?  1/ 

4.    Trace  the  curve-1-  +tt  =  1« 

The  curve  cuts  the  axis  of  x  at  (a,  0)  and  the  axis  of  y  at 
(0,  b).     There  is  one  infinite  branch  in  the  second   angle  and 

another  in  the  fourth,     y  = a?  is  the  equation  of  the  asymp- 


tote, which  lies  below  the  infinite  branches, 
cave  upward,  except  between  x  =  0  and 
x  =  a,  where    it   is    concave  downward. 
(0,  b)  and  (a,  0)  are  points  of  inflexion. 

5.    Trace  the  curve  y  = 


The  curve  is  con- 


1+x2 

The  curve  has  one  infinite  branch  in  the 
first  angle  and  another  in  the  third,  to  each 
of  which  the  axis  of  x  is  an  asymptote. 
}  is  a  maximum,  and  —  %  a  minimum, 
ordinate.  (0,  0),  (—  V3,  —  £V3),  and 
(V3,  iV3)  are  points  of  inflexion.  The 
inclination  of  the  curve  at  the  origin  is  £71-. 


Fig.  46. 


6 .    Trace  the  curve  y2  =  x2 
see  Fig.  46. 


2a  -\-x 


For  the  form  of  the  curve, 


168  CURVE  TRACING. 

151.  Tracing  Polar  Curves.  When  possible,  write  the  equa- 
tion in  the  form  p  =f(0) .  Solve  f{6)  =  0  to  find  the  angles  at 
which  the  curve  cuts  the  polar  axis  at  the  pole.  Assign  to  6 
such  positive  and   negative  values   as   make   p  easily  found. 

Solve  -£  =  0  to  find  the  values  of  0  for  which  p  is  a  maximum 

or  minimum,  and  for  which  the  curve  is  perpendicular  to  the 
radius  vector.  Examine  the  curve  for  asymptotes,  direction  of 
curvature,  and  points  of  inflexion.  The  facts  thus  obtained  will 
indicate  the  form  of  the  curve. 


Examples. 

1.  Trace  the  curve  p  =  a  sin  3  9. 

Since  p  =  a  sin  30,  p  reaches  its  maximum  value  a  when 
sin 3  0=  1  ;  that  is,  when  0  =  171-,  f  tt,  -§7r,  etc. ;  and  p  reaches 
its  minimum  —  a  when  sin  3d  =  —  1 ;  that  is,  when  $  =  |-7r,  ^7r, 
JJ-7T,  etc. 

Since  -^  =  3  a  cos  30,  p  increases  from  0  to  a,  while  6  in- 
creases from  0  to  ^7r ;  p  decreases  from  a  to 
—  a,  while  6  increases  from  ±tt  to  £71-;  p 
increases  from  —  a  to  +0,  while  6  increases 
from  \tr  to  f  tt  ;  and  p  decreases  from  a  to 
0,  while  6  increases  from  fir  to  tt.  Further 
revolution  of  the  radius  vector  in  either 
Fig-  47-  direction  would  evidently  retrace  the  three 

loops  already  found.     The  curve  is  that  represented  in  Fig.  47. 

2.  Trace  the  curve  p  =  a  sin  20. 

The  curve  consists  of  four  loops.  From  this  and  the  previous 
example,  we  infer  that  the  locus  of  p  =  asinnO  consists  of  n 
loops  when  n  is  odd,  and  2n  loops  when  n  is  even. 

8  n 

3.  Trace  p  =  a  sin  —  4.    Trace  the  lituus  p  = 

5.   Trace  p  =  a.  cos  6  +  6,  in  which  a  >  b. 


miscellaneous  examples.  1g9 

Miscellaneous  Examples. 

1.  Examine  a4  —  axy2  —  cof  =  0  for  multiple  points. 

The  origin,  a  triple  point,  is  a  cusp  of  the  first  species,  through 
which  a  branch  of  the  curve  passes. 

2.  Examine  ax5  +  by3  —  c  =  0  for  points  of  inflexion. 

3.  Prove  that  (0,  0)  is  a  multiple  point  of  intersection  on  the 
curve  a*4  —  a2xy  -f-  b2y2  =  0. 

4.  Examine  x*  —  arx2  +  d*y  =  0  for  points  of  inflexion. 

Ans.     — — ,  —   ,    and -,  —  )• 

W(3    36/  \     V6    36/ 

5.  Examine  ay2  —  x*  —  bx2  =  0  for  multiple  points. 

6.  Examine  ay2  —  x3  +  bx2  =  0  for  multiple  points. 

Ans.    (0,  0)  is  a  conjugate  point. 

tyy  q3 

7.  Trace  the  curve  y-  = 

x-\-  b 

8.  Trace  the  folium  of  Descartes  y*  —  3  axy  -f  x*  =  0. 

9.  Trace  (y  —  x2)2  =  x5. 

10.  Trace  y2(x  -  a)  =  x-3. 

1 1 .  Trace  p  cos$  =  a  cos  26. 


CHAPTER   XII. 


CURVATURE,  EVOLUTES,  ENVELOPES,  AND  ORDER  OF 
CONTACT. 

152.    If  (f>  represent  the  inclination  of  any  curve  ab  referred 
to  rectangular  axes,  then  <f>  will  measure  the  direction  of  the 
fj  curve  with  respect  to  the  axis  of  x.     At  the 

point  p,  4>  =  angle  xor  ;  and  at  p',  c/>  =  xmp'  ; 
hence  angle  prm  =  Ad>,  if  arc  pp'=  As,  s  repre- 
senting the  length  of  the  curve. 


153.  The  Curvature  of  a  curve  at  any  point 
is  the  rate  of  change  of  its  direction  relative 
to  that  of  its  length. 


Fig.  48. 


Hence,  if  k  represent  the  curvature  of  any  curve, 


K  =  #,  or   J"* 

ds         As  =  0 


"A  (ft" 
As 


§31. 


154.  Curvature  of  a  Circle.  If  ab  (Fig.  48)  be  the  arc  of  a 
circle  whose  radius  is  r,  the  angle  prm  equals  the  angle  sub- 
tended by  the  arc  pp'  at  its  centre ;  and  from  §  40  we  have 


angle  prm  = 


arc  pp 


Ad>      1 

or  — ^  =  - 

As      r 


d$ 
ds' 


or  k,  =  -  ■ 
r 


§  17,  Cor.  2. 


Hence,  the  curvature  of  any  circle  is  equal  to  the  reciprocal  of 
its  radius;  and  the  curvatures  of  any  two  circles  are  inversely 
•proportional  to  their  radii. 

Cor.  If  r=l,  k=  1 ;  that  is,  the  unit  of  curvature  is  the  cur- 
vature of  a  circle  whose  radius  is  unity. 


RADIUS   OF   CURVATURE. 


171 


155.    To  find  k  in  terms  of  the  differentials  o/x  and  y. 

tan  <&  =  —  ; 
v      dx 

o,  7 ,       d-y 
dx 


,:dcf> 


dx 


1  + 


ds 


2,  since  sec2<£  =  l  -f 

djt 
dx- 


dx) 


1  + 


We  take  the  positive  value  of  the  radical  so  that  the  curva- 
ture of  a  curve  will  be  positive  or  negative  according  as  — ^  is 

positive  or  negative  ;  that  is,  according  as  the  curve  is  concave 
upward  or  downward.  The  sign  of  curvature,  however,  is  often 
neglected. 


156.  Radius  of  Curvature.  As  the  radius  of  a  circle  varies 
from  0  to  oo,  its  curvature  varies  from  co  to  0  ;  hence  there  is 
always  a  circle  whose  curvature  is  equal  to  that  of  an}'  curve  at 
any  point.  A  circle  tangent  to  a  curve,  and  having  the  same 
curvature  as  the  curve  at  the  point  of  contact,  is  called  the 
Circle  of  Curvature  of  the  curve  at  that  point.  Its  radius  and 
centre  are  the  radius  of  curvature  and  the  centre  of  curvature  of 
the  curve  at  that  point.  Hence,  if  R  represent  the  radius  of 
curvature  of  a  curve,  from  §§154  and  155,  we  have 

3 


B  =  -  = 


Wl 


dx- 


R  will  be  positive  or  negative,  according  as  the  curve  is  con- 
cave upward  or  downward  ;  but  its  sign  is  often  neglected. 


172  CURVATURE. 

157.  The  radius  of  curvature  in  terms  of  polar  coordinates 
can  be  found  by  transforming  the  value  of  M  in  §  156  to  polar 
coordinates.     "We  thus  obtain 

djf^ 


.  0  dp-        d'p        2  i   a  do1         cVp 

p-  _f_  V  — a  — '-       p    -f-  I —  p ; 

1  dO-      '  dd-  d?     '  d6- 

in  which  JVis  the  normal.     See  §  131,  Ex.  8,  and  §  136,  (7). 

158.  The  circle  of  curvature,  in  gerieral,  cuts  the'  curve  at  the 
point  of  contact. 

For,  on  one  side  of  the  point  of  contact,  the  curve  changes 
its  direction  more  rapidly  than  the  circle  of  curvature,  and  hence 
lies  within  the  circle  ;  while  on  the  other  side  it  changes  its 
direction  more  slowly  than  the  circle,  and  hence  lies  without  the 
circle. 

159.  At  a  point  of  maximum  or  minimum  curvahire,*  the  cir- 
cle of  curvature  does  not  cut  the  curve;  and  conversely. 

For,  on  either  side  of  a  point  of  maximum  curvature,  the  curve 
changes  its  direction  more  slowly  than  at  this  point ;  hence,  on 
each  side  of  this  point,  the  curve  lies  without  the  circle  of  curva- 
ture at  this  point.  On  either  side  of  a  point  of  minimum 
curvature,  the  curve  changes  its  direction  more  rapidly  than  at 
this  point ;  hence,  on  each  side  of  this  point,  the  curve  lies 
within  the  circle  of  curvature. 

Since  the  conic  sections  are  symmetrical  with  respect  to  nor- 
mals at  their  vertices,  it  follows  that  their  vertices  must  be  points 
of  maximum  or  minimum  curvature. 

Examples. 
1 .    Find  the  curvature  of  the  parabola  y2  =  2px. 


Here        -f-  =*- ,  aud  — %  =  —  ^  ; 


cly^ 

dx      y  dx2  y3 


*  By  a  maximum  or  minimum  curvature,  we  mean  a  numerical  maxi- 
mum or  minimum,  the  sign  of  curvature  not  being  considered. 


EXAMPLES.  173 

dh/ 

dx2 V  (    y-    V_  ±        ir 


l+®j 


tfW+w        itf+i?)* 


The  upper  or  lower  sign  is  to  be  taken  according  as  —  ^j  is 
positive  or  negative. 

At  the  vertex  (0,  0),  *  =  -,  which  is  evidently  the  maximum 
curvature  of  the  parabola. 

2.    Neglecting   its    sign,  find   the   curvature   of   the   ellipse 

a2y2  +  Ifx-  —  crb2. 

Here         &  =  -5?,  and  ^  =  -  -^ ■; 

dx  uJ>/  dxr  cryi 

V  /      a4?/2      \i  a*b- 


ccif  \a4y2  +  b*x*J       (aAy2  +  &4ar)  ? 

At  the  vertex  (a,  0) ,  k  =  —  ;   at  the  vertex  (0,  —J) ,  *  =  —  ; 

b~  (i 

hence,  the  maximum  curvature  of  the  ellipse  is  — ,   and   the 

.   .  b  b~ 

minimum—- 


3.    Find  the  radius  of  curvature  of  the  cvcloid 


a;  =  r  vers    -  —  -\/2ry  —  y2. 


TT  dv      V'2  ry  —  if         ,  d2y  r 

Here        —  = 2 — -,  and  -r^  = ?5 

dx  y  dx-  y 

.•.a=(f)'K)~^ 

which  equals  numeric-all}-  twice  the  normal. 

1 

k  = -=; 

-2V'2ry 

and ,  the  curvature  at  the  highest  point,  is  evidently  the 

4?- 
minimum  curvature  of  the  cycloid. 


174  EVOLUTES. 

4.  Find  the  curvature  of  y  =  xi—  4^  —  18a^  at  the  origin. 
Find  the  abscissas  of  the  points  at  which  the  curvature  is  0. 

Ans.    k  =  —  36  ;  x  =  3  and  —  1 . 

5.  Find  the  curvature  of  the  logarithmic  curve  y  =  ax. 

Ans.   k  =  —  1  ■ ''   — 
(mz  +  2r)s 

6.  Find  the  numerical  value  of  the  radius  of  curvature  of  the 

cubical  parabola  y3  =  a2x.  (qv*_l.  *\z 

Ans.    R  =  LJLX_2_. 
Ga'y 

7.  Find  the  radius  of  curvature  of  the  spiral  of  Archimedes 
p  =  ad. 

Here        —  =  a,    and   — ~  =  0 ; 
dQ  d&2 

02  +  2^-0^       /32  +  2a2  2+02 

8.  Find  the  radius  of  curvature  of  the  logarithmic  spiral 
p~     '  Ans.  B  =  pVl  +  (log a)2. 

B  volutes. 

160.  The  Evolute  of  a  given  curve  is  the  locus  of  the  centre 
of  curvature  of  the  curve.  The  given  curve  is  called  the 
Involute  of  its  evolute. 

161.  To  find  the  equation  of  the  evolute  of  any  given  curve. 
Let  c,  (a,  /?) ,  be  the  centre  of  curvature  of  the  curve  ab  at  any 

point  p,  (x,  y).     Then, 

since  bp  =  pc  sin  bcp  =  R  sin  kph  =  R — , 

ds 

and  bc  =  i?  cos  bcp  =  R— -  ; 

ds 

a=  OE  —  BP  =  X—  R  — ,      (1) 

ds      v  ' 
and         /?  =  ep  +  bc  =  y-j-R—-    (2) 


EXAMPLES. 


175 


Substituting  in  (1)  and  (2)  the  values  of  R  and  ds,  we  have 

(3) 


a  =  X 


1  +  f!£\dl 

da?)  dx 


1  + 


and 


P  =  y  + 


dry 

dH? 

dtf 


&y 

dx2 


(4) 


By  differentiating  the  equation  of  any  given  curve,  and  sub- 
stituting the  results  in  (3)  and  (4),  a  and  /5  may  be  expressed 
in  terms  of  x  and  y.  If,  between  the  equations  thus  obtained 
and  that  of  the  given  curve,  x  and  y  be  eliminated,  the  resulting 
equation  between  a  and  /3  will  be  the  equation  of  the  evolute. 


Examples. 

1.    Find  the  equation  of  the  evolute  of  the  parabola. 

Here        ^.—L,    and  _^  =  _^. 
dx     y  dx-  y* 

Substituting  these  values  in  (3)  and  (4)  of  §  161,  and  reducing, 
we  have 


and 


a  =  3  x  +_p,   or  x  =      t     , 
o 

V3 

/?  =  -^,,  or  y  =  -fihpi. 


Substituting  these  values  of  x  and  y 

in   the   equation    of  the    parabola,    we   Q 

have 

/3^  =  |p(a-p), 


or 


*-s?  <—*>•• 


which  is  the  equation  of  the  evolute  of 

»        a  Fie.  50 

y-  =  2p.c.  B 

bac  (Fig.  oO)  is  the  evolute  of  the  parabola  mon. 


176 


EVOLUTES. 


2.   Find  the  equation  of  the  e volute  of  the  ellipse. 

&  —  &   an 

\x         a-y 

_  (a2-b2)a? 


TT  dy         b2x         ,  dry  &4 

Here        -£  =  —  —,  and  -2  =  — _; 
die         cry  da?         a-y 


-,  or  x  = 


a  a 
a2  —  61 


and 


f (X    ~~  I)    ill 

(3  =  -  ± — ir^-'  or 


V«2-62; 


Substituting  these  values  of  x  and  ?/  in  a2?/2  -(-  62#2  =  a262,  we 
obtain  as  the  equation  of  the  evolute  of  the  ellipse, 

(aa)  i +  (&/?)  t  =  (a2 -62)f. 


3.    Find  the  equation  of  the  evolute  of  the  cycloid 

x  =  r  vers-1  — —  V2  ry  —  y2. 
Here        ^  =  V2ry-y'>  and  <g  =  _  r_. 


(1) 


dx 


dx2 


y 


.-.y  =  -(3,  and  x=  a^  2V —  2r/2  - /32. 

Substituting  these  values  of  #  and  y  in  (1),  we  obtain  as  the 
equation  of  the  evolute  of  the  cycloid, 

a  =  r  vers" 1  f-  &\  +  V-  2  r/3  -  /32.  (2) 

The  locus  of  (2)  is  another  cycloid  equal  to  the  given  cycloid, 

the  highest  point  being 
at  the  origin.  For,  if 
1  the  cycloid  oOjM  be 
referred  to  the  axes 
OxXj  and  o^,  and  p  be 
any  point, 

y  =  -  bp, 

and  x  =  0{B 

=  ke  +  dp.    (3) 

em  =  arc  ep  : 


PROPERTIES  OF  THE  EVOLUTE.  177 


.-.  ke  =  arc  ap  =  r  vers  1  —  =  r  vers  x  (  — -  )•  (4) 


dp  =  Vad-de  =  V—  y('2r  +  y)'=  V—  2  ry  —  y-.  (5) 
From  (3),  (4),  and  (5),  we  obtain 

x  =  r  vers  x  f—j  +  V-2ry->/-.  (0) 

Comparing  (2)  with  (6),  we  see  that  08,  the  evolute  of  oo1? 
must  be  equal  to  opi ;  that  is,  the  evolute  of  a  cycloid  is  an  equal 
cycloid. 

Properties  of  the  Evolute. 

162.    Any  normal  to  the  involute  is  a  tangent  to  the  evolute. 
The  equation  of  the  normal  to  y  =  f(x)  at  (x1,  y')  is 
rtV 

y-y  '=--7-7(3?-*').  (i) 

dy' 

Let  (a,  /?)  be  the  centre  of  curvature  of  y  =  f(x)  at  (V,  y') ; 
then  (1)  passes  through  (a,  /?),  and  we  have 

y-/?  =  "f?  (X'-a)]  (2) 

.:x'-a  +  (^(y'-(3)=0.  (3) 

If  (#',  y')  move  along  the  involute,  (a,  /?)  will  move  along 
the  evolute,  and  a,  (3,  and  y'  will  be  functions  of  x'.  Differen- 
tiating (3)  on  this  hypothesis,  we  have 

ax'  dx' 

Dividing  by  dx',  and  rearranging  terms,  we  have 

l  +  ^2  +  (v'      m^'-^-^=0.  (4) 


178  EVOLUTES. 


But,  as  (a,  /S)  is  on  the  evolute,  we  have  by  §  161,  (4), 
1  + 


dyn 


dx'2 

f-f'  +  Tgr-; 

dx'2 


,.0,_/0gg+1+«g_o.  (5) 

From  (4)  and  (5),  we  obtain 

_  da      dy'd/3  _  Q       .   _da'  =  d/?# 
da;'       da;'2  d?/'      da 

Hence  (1)  is  tangent  to  the  evolute  at  (a,  /?). 

163.  Any  continuous  arc  of  the  evolute  is  equal  to  the  differ- 
ence between  the  radii  of  curvature  of  the  involute  that  are  tangent 
to  (his  arc  at  its  extremities. 

If  (x  —  a)2  +  (y  —  /3)2  =  Er  be  the  circle  of  curvature  of  any 
curve  at  (a;',  y') ,  we  have 

(x'-ay+(tf-py=ii*.  (i) 

Suppose  (x',  y')  to  move  along  the  curve  ;  then  y\  a,  /?,  and 
B  will  be  functions  of  x'. 

Differentiating  (1)  on  this  hypothesis,  we  obtain 
(x'—a)dx'  +  (y'—p)dy'—(x'—a)da. 

-(y'-P)dp  =  RdM.  (2) 

From  equation  (2)  of  §  162  we  have 

dy' 

,.y'-(3  =  (^(x'-a).  (4) 

From  (3), 

(x'-a)dx'+(y'-p)dy'=0.  (5) 

From  (2)  and  (5), 

(x'-a)da  +  (y'-(3)d(3  =  -RdR.  (6) 


PHOPEItTIES    OF    THE    EVOLUTE. 


179 


From  (1)  and  (4), 


V  '  dar 


(7) 


From  (4)  and  (G), 


(8) 


da 
Squaring  (8)  and  dividing  by  (7),  we  obtain 

ck2  +  d/32  =  c7222. 
But,  s  being  the  length  of  the  e volute,  we  have 
da2  +  d/32  =  cZs2 ; 
.-.  ds  =  ±  (?i? ; 
that  is,  R  increases  or  decreases  as  fast  as  s  increases. 

Hence,  if  the  length  of  the  evolute  of  the  parabola  (Fig.  52) 
be  estimated  from  the  point  a,  we  have 

arc  ap  =  kp  —  oa. 

Again,  if  the  length  of  the  evolute  of  the  cycloid  (Fig.  51) 
be  estimated  from  o, 

B 

arc  os  =  soj  =  4  r ; 

hence  the  length  of  one  branch  of  the 
cj'cloid  is  8  r. 

164.  These  two  properties  of  the 
evolute  enable  us  to  regard  any  invo- 
lute as  traced  by  a  point  in  a  string 
unwound  from  its  evolute.  Thus,  if 
on  a  pattern  of  cpao  one  end  of  a 
string  be  fastened  at  c,  and  the  string 
be  then  stretched  along  the  right  of 
cpa,  the  point  of  the  string  which 
reaches  o,  when  carried  around  to  the  right,  will  trace  the  arc 
okwi  as  the  string  unwinds  from  the  evolute. 


Fig.  52. 


180 


ENVELOPES. 


Since  an}-  point  of  the  string  beyond  a  will  trace  an  involute 
of  ca,  it  follows  that,  while  a  curve  has  but  one  evolute,  it  can 
have  an  infinite  number  of  involutes. 

Envelopes. 

165.  If,  in  the  equation  f(x,  y,  a)  =  0,  a  series  of  different 
values  be  assigned  to  a,  the  equation  will  represent  a  series  of 

curves  differing  in  form,  or  in 
position,  or  in  both  these  re- 
spects, but  all  belonging  to 
the  same  class  or  family  of 
curves. 

For  example,  if  different 
values  be  assigned  to  a  in 
(x  —  a)2  +  y-  —  25,  its  loci  will 
be  a  series  of  equal  circles  with  their  centres  on  the  axis  of  x 
(Fig.  53). 

The  quantity  a,  which  is  constant  for  any  one  curve,  but 
changes  in  passing  from  one  curve  to  another,  is  called  a  Vari- 
able Parameter.  Any  two  curves  of  a  series  that  correspond 
to  nearly  equal  values  of  the  parameter  usually  intersect,  and 
are  called  consecutive  curves. 


Fig.  53. 


166.  An  Envelope  is  the  locus  of  the  limiting  positions  of  the 
points  of  intersection  of  the  consecutive  curves  of  a  series,  as 
these  curves  approach  indefinitely  near  each  other. 

167.  To  find  the  equation  of  the  envelope  of  a  series  of  curves. 
Let  f(x,  y,  a)  =  u  =  0  (1) 

and  /(a?,  y,  a  +  Aa)  =  0  (2) 

be  the  equations  of  any  two  consecutive  curves  of  a  series  ;  then, 
at  the  points  of  intersection  of  (1)  and  (2),  we  evidently  have 


f(x,  ?/,  a  +  Aa)  -f(x,  y,  a)  _  Q 
Aa 


(3) 


ENVELOPES.  181 

Passing  to  the  limit  as  Aa  =  0,  we  have,  at  the  limiting  posi- 
tions of  these  intersections, 

-f/(a>,y,a)=0,  or  ^  =  0.  (4) 

da  (fa 

Since  the  coordinates  of  the  points  on  the  envelope  satisfy 
both  (4)  and  (1),  its  equation  is  found  by  eliminating  u  between 
these  equations. 

168.    The  envelope  is  tangent  to  each  curve  of  the  series. 
Let  <f>  (x,  y)  represent  the  value  of  a  obtained  from   (4)  of 
§  1G7  ;  then,  if  a  =  cj>  (x,  y), 

f(x,y,a)  =  it=0  (1) 

is  the  equation  of  the  envelope. 

Differentiating  (1),  a  being  variable,  we  obtain 

du  7     .  clu  7     .  du,        r. 

—  ax  -i dy  -\ «a  =  0. 

dx  dy  da 

But,  at  any  point  on  the  envelope, 

ff  =  °;  §167,(4) 

aa 

.:*Hdx  +—dy  =  0,  (2) 

dx  dy 

which  gives  the  slope  of  the  envelope  at  an}-  point. 

Differentiating  f(x,  y,  a)  =  u  =  0,  considering  a  constant,  we 
have  7  7 

^dx  +  ^dy^O,  (3) 

dx  dy 

which  gives  the  slope  of  any  individual  curve  at  am*  point. 
From  (2)  and  (3)  we  see  that  the  envelope  and  any  curve  of 
the  series  have  the  same  slope  at  their  common  point. 

Examples. 

1.  Find  the  envelope  of  (x  —  a)2-\-y2  =  r2,  in  which  a  is  a 
variable  parameter. 


182  ENVELOPES. 

Here       f(x,  y,  a)  [=  u]  =  (x  -  a)2  +  f  -  r2  =  0  ;  (1) 

.:^  =  -2(x-a)  =  0.  (2) 

From  (1)  and  (2) ,  y  =  ±  r ;  that  is,  the  envelope  is  two  lines 
parallel  to  the  axis  of  &,  as  would  be  inferred  from  Fig.  53. 

2.  Find  the  envelope  of  y  =  ax+  — ,  a  being  the  variable 
parameter. 

Here        /(a?,  ?/,  a)[=  w]  =  y  —  aa— ™  =  0  ;  (1) 

.\— =  — as  +  — =  0.  (2) 

Cta  a 

Eliminating  a  between  (1)  and  (2),  we  obtain  y2  =  imx  ;  that 
is,  the  envelope  is  a  parabola  whose  latus  rectum  is  4m. 

3.  Find  the  envelope  of  the  hypotenuse  of  a  right-angled 
triangle  of  constant  area. 

Let  a  and  /5  be  the  sides  of  the  right  triangle,  and  assume 
them  as  coordinate  axes  ;  then 

*  +  ^  =  l 

a       (3 

is  the  equation  of  the  hypotenuse. 
Let  c  =  the  constant  area  ;  then 

a/3  =  2  c,  or  /?  =  — 

a 

Hence     f(x,  y,  a)  =  *  +  ^-  1  =  0,  (1) 

a       2c 

and  ^  =  _-^-  +  J-=:o.  (2) 

da  a2      2c  v  ' 

Eliminating  a  between  (1)  and  (2),  we  obtain  xy  =  \c,  that 
is,  the  envelope  is  an  hyperbola  to  which  the  sides  of  the  tri- 
angle are  asymptotes. 


CONTACT   OF   DIFFERENT   ORDERS.  183 

4.  Find  the  envelope  of  a  system  of  concentric  ellipses,  the 
area  and  the  directions  of  the  axes  being  constant. 

Let  the  equation  of  the  ellipses  be 

aY  +  (32X-  =  a2/3-,  (1) 

and  c  represent  the  constant  area  ; 
then  c  =  Tra(3.  (2) 

Eliminating  ft  between  (1)  and  (2),  we  have 

J\x,  y,  a)  =  ay +  4^-  4  =  0  ;  (3) 

TT'a"  TV" 

/.^=2a3/2-^f=0.  (4) 

Ota  7r"a 

From  (3)  and  (4),  we  obtain 
c 

which  are   the  equations  of   conjugate  equilateral  hyperbolas 
referred  to  their  asymptotes. 

169.  Contact  of  Different  Orders.  Let  y=f{x)  and  y  =  4>(x) 
be  any  two  curves  referred  to  the  same  axes.  If/(a)  =  </>(«), 
the  curves  have  the  point  [«,/(«)]  in  common.  If  /(«)  =  <f>(a) 
and  f'(a)=  <f>'(a),  the  curves  are  tangent  at  [a,/(a)],  and  are 
said  to  have  a  contact  of  the  first  order.  If  /(«)  =  </>(a),  /'(a) 
=  <£'(a),  and  /"(a)  =  <£"(a),  the  two  curves  have  the  same 
curvature  at  their  common  point,  and  their  contact  is  of  the  sec- 
ond order.  If  in  addition,  f'"(a)=  </>'"(a),  their  contact  is  of 
the  third  order;  and  so  on.  Thus,  contact  of  the  nth  order  im- 
poses n  +  1  conditions. 

170.  Two  curves  intersect  or  do  not  intersect  at  their  point  of 
contact,  according  as  their  order  of  contact  is  even  or  odd. 

Let  y—f(x)  and  y  =  <f>(x)  be  any  two  curves  having  the  point 
[«,/(«)]  in  common.  Let  h  be  a  very  small  increment  of  x. 
By  Taylor's  formula,  we  have 


184  ORDEIi   OF   CONTACT. 

/(a  +  h)  =/(a)  +f{a)h  +/"(a) -|  +/'"(«)  *  +  ...     (1) 

^(a  +  fe)  =  ^(a)+^(a)A  +  ^'(a)-^  +  ^"(a)-^  +  ...(2) 

Subtracting  (2)  from  (1),  we  obtain 
/(a+A)-^(a+A)=fe[/'(a)-^(a)]+-^[/»(a)-^(a)3 

+  |f  [/'"(«)-<^'»]  +  jj-[/IV(«)-^(a)]+-,  (3) 

whicb  gives  the  difference  between  the  corresponding  ordinates 
of  the  curves  on  each  side  of  their  common  ordinate.  If  the 
contact  is  of  an  odd  order,  the  first  term  of  the  second  member 
of  (3)  which  does  not  vanish  contains  an  even  power  of  h ; 
hence  the  sign  of  the  second  member  is  the  same  whether  h  be 
positive  or  negative.  Therefore,  y=f(x)  lies  either  above  or 
below  y  =  <£ (x)  on  both  sides  of  their  common  point;  and  the 
curves  do  not  intersect.  But,  if  their  contact  is  of  an  even 
order,  the  first  term  of  the  second  member  of  (3)  that  does  not 
vanish  contains  an  odd  power  of  k.  Hence,  in  this  case,  the 
second  member  changes  sign  with  It ;  and  y=f(x)  lies  above 
y  =  <f>(x)  on  one  side  of  their  common  point,  and  below  it  on  the 
other ;  and  the  curves  intersect. 

Cor.  At  a  point  of  maximum  or  minimum  curvature,  the 
circle  of  curvature  has  contact  of  the  third  order  with  the  curve  ; 
for  it  does  not  cut  the  curve  at  such  a  point. 

The  following  are  obvious  conclusions  from  equation  (3)  : 

(a)  Two  curves  on  each  side  of  their  common  point  are 
nearer,  the  higher  their  order  of  contact. 

(b)  If  two  curves  have  a  contact  of  the  wth  order,  no  curve 
having  with  either  of  them  a  contact  of  a  lower  order  can  lie 
between  them  near  their  common  point. 

171.  Osculating  Curves.  The  curve  of  a  given  species,  that  has 
the  highest  order  of  contact  possible  with  a  given  curve  at  any 
point,  is  called  the  osculating  curve  of  that  species. 


OSCULATING  CURVES.  185 

Let  y  =f{x)  be  the  most  general  form  of  the  equation  of  a 
curve  of  a  given  species,  and  suppose  that  it  contains  n-\-l 
arbitrary  constants.  Upon  the  n  -f  1  constants,  n  +1,  and  only 
w+1,  independent  conditions  can  be  imposed.  But,  in  order 
that  y  =f{x)  may  have  contact  of  the  nth  order  with  a  given 
curve  at  a  given  point,  n  4-1  conditions  must  be  fulfilled  by  its 
constants  ;  that  is,  these  constants  must  have  such  values  that 
y=f(x)  shall  pass  through  the  point,  and  the  first  n  derivatives 
of  its  ordinate  be  equal  to  those  of  the  given  curve  at  this  point 
(§  169). 

Hence,  as  y  =  ax  +  b  has  two  constants,  the  osculating  straight 
line  has  contact  of  the  first  order,  and  is  a  tangent. 

As  (x  —  a)2  +  (y  —  b)2  =  r2  has  three  constants,  the  oscidating 
circle  has,  in  general,  contact  of  the  second  order,  and  is  the  circle 
of  curvature. 

The  osculating  parabola  has  contact  of  the  third  order.  The 
osculating  ellipse  or  hyperbola  has  contact  of  the  fourth  order. 

Miscellaneous  Examples. 

1.    Find  the  curvature  of  the  hyperbola. 

a  a*b* 

Ans.  k  =  ± 


(ftW  +  oV)l 

2.  Find  the  radius  of  curvature  of  the  equilateral  hyperbola 
xy  =  \a\  Ans   B=s±(*?+y)im 

a- 

3.  Find  the  radius  of  curvature  of  if  =  Gar  +  ar\ 

Ans.  R  =  J-2 — — » ! '— L  numerically. 

4.  Find  the  radius  of  curvature  of  the  lemniscate  of  Ber- 
noulli, p2=  a2  cos  26.  ,        „  _  a2 

5.  Find  the  curvature  of  the  cissoid  y  = 


2  a  —  x 

Ans.    «  =  ±  3<2g-»y 

oa£(8a  —  3x)i 


186  INVOLUTES  AND  ENVELOPES. 

G.    Find  the  equation  of  the  evolute  of  the  hyperbola. 

Ans.    (aa)  1  -  (b(3)  I  =  (a2  +  b2)  §. 

7.    Find  the  length  of  the  evolute  of  the  parabola  in  terms  of 
the  abscissas  of  its  extremities. 

B  =  (y2+r)^(2x+P)^  §159,  Ex.1. 

Arc  ap  (Fig.  52)  =  kp  —  oa=  -^ -^-  —  p 


if-fj-^  »"w- 


8.  Find   the   envelope   of   y2  —  a.  (x  —  a) ,   in  which   a   is   a 
variable  parameter. 

Ans.    y  =  ±  -J-a;. 

9.  One  angle  of  a  triangle  is  constant  and  fixed  in  position  ; 
find  the  envelope  of  the  opposite  side  when  the  area  is  constant. 

Ans.   xy  = ,  c  being  the  constant  area  and  w  the  con- 

2sinw 

stant  angle. 

10.  Find  the  envelope  of  the  circles  whose  diameters  are  the 
double  ordinates  of  the  parabola  y2  =  2px. 

Ans.   y2  =p  (p  +  2  x) . 

11.  Find  the  envelope  of  a  line  of  constant  length  a,  whose 
extremities  move  along  two  fixed  rectangular  axes. 

Ans.   x*  4-  y%  =  a*. 


12.  Find  the  envelope  of  the  normals  to  the  parabola  y2=2px. 

o 

Ans.   y2  = (x  —  pY,  which  is  the  evolute  of  the  parabola, 

27p 

as  it  clear  1}T  should  be. 

13.  Find  the  radius  of  curvature  of  the  catenary  2/=— f  e«  _f_  e  «  ). 

Ans.    E  =  — 


MISCELLANEOUS    EXAMPLES.  187 

14.  Find  the  radius  of  curvature  of  the  cardioid  p=a{\  —  cos0). 

(Hap)} 
3 


Ahs.    R  = 


15.  Find  the  envelope  of  the  series  of  ellipses  —  -\ £ =  1 , 

a2     (k—a)2 
a  being  a  variable  parameter. 

Ans.    ajl  -f  y^  =  A-i. 

16.  Prove  that  the  whole  length  of  the  evolute  of  the  ellipse 

.     .  a'1  -  bz 

is  1 

ab 


CHAPTER   XIII. 
INTEGRATION  OF  RATIONAL  FRACTIONS. 

172.  Decomposition  of  Rational  Fractions.  Any  rational  frac- 
tion whose  numerator  is  not  of  a  lower  degree  than  its  denomi- 
nator can  be  separated  by  division  into  two  parts,  the  one 
rational  and  entire,  and  the  other  a  rational  fraction  whose 
numerator  is  of  a  lower  degree  than  its  denominator.  For 
example, 

xA  0  .  5  xr  —  4 

■ — —  =  x  —  2  H 

ar  -f-  2  ar  —  x  —  2  ar  +  2  a-  —  x  —  2 

Hence  any  rational  differential  can  be  considered  as  com- 
posed of  an  entire  part  and  a  fraction  whose  numerator  is  of  a 
lower  degree  than  its  denominator.  The  entire  part  can  be 
integrated  by  previous  methods  ;  and  it  is  our  present  object  to 
show  that  the  fractional  part,  if  not  directly  integrable,  can  be 
resolved  into  partial  fractions  which  are  integrable.  These  par- 
tial fractions  differ  in  form,  according  as  the  simple  factors  of 
the  denominator  of  the  given  fraction  are  : 

I.  Real  and  unequal. 

II.  Real  and  some  of  them  equal. 

III.  Imaginary  and  unequal. 

IV.  Imaginary  and  some  of  them  equal. 

To  show,  in  the  simplest  manner,  how  the  decomposition  and 
integration  is  to  be  effected,  we  shall  apply  the  process  to  par- 
ticular examples  in  each  of  the  four  cases. 

173.  Case  I.  When  the  simple  factors  of  the  denominator  are 
real  and  unequal,  to  every  factor,  as  x  —  a,  there  corresponds  a 

partial  fraction  of  the  form 

x  —  a 


CASE   T.  189 


Let  it  be  required  to  find  |  ^   '  "*~ — ^ 

J  x3  +  af  —  2  x 


+  : 

The  roots  of  x3  -\-xr—2x  =  0  are  0,  1,  and  —2;  hence,  the 
factors  of  a?  +  or  —  2  x  are  a;,  x  —  1 ,  and  a;  4  2. 

A  2a; +  3  A  ,'B      ,      0  /1N 

Assume  ! = ( l ) 

x(x  —  l)(a?4-2)       a?      a?— la? +  2  v 

Clearing  (1)  of  fractions,  we  have 

2  a;  +  3  =  A(x—l)(x  +  2)  +  5  (a:  4-  2)a? 

4-C(aj-l)as  (2) 

=  (A+B+C)x2+(A+2B-C)x-2A.   (3) 
Equating  the  coefficients  of  like  powers  of  x  in  (3),  we  have 
A+B+C=Q>,  A+2B-C=2,  and  -2^1=3.   (4) 
Solving  equations  (4),  we  find 

A  =  -%,  B  =  |,  andC  =  -f  (5) 

Substituting  these  values  in  (1),  we  obtain 

2^4-3  3  5  1  ,„,. 


ar*  4-ar!-2a;  2a;     3(a;-l)      6(a?  +  2)' 

.    r(2x_+_3)dx  _  _  3  Ato      5  f  da;    _  1  /•  da; 

"J  afJ  +  ar  — 2a;  2J  a;       3J  a;  — 1      6 J  i 


4-af  — 2aj  2  J  a;       3J  x-1      6»/a;4-2 

=  -  f  log  a?  4- f  log(a>  - 1)  -  £  log  (a;  +  2)  4-  logc 

=  logc^'-1>5. 
sa;B(a;4-2)& 

Equation  (G)  is  true  ;  for  the  values  of  A,  B,  and  C  given  in 
(5)  satisfy  (4),  and  hence  make  (3),*  and  therefore  (1),  an 
identical  equation. 

*  The  principle  used  here  is:  I/A  =  Af,  B  =  B',  C  =  C,  etc.,  A  +  Bx 
4  Cx2  +  •••  =  A'+  B'x  4  C'x2  4  •••  is  an  identical  equation. 

That  the  fraction  has  been  correctly  decomposed  is  proved  by  reasoning 
backward  through  the  process.  Equation  (1)  is  not  assumed  as  a  basis  of 
proof,  but  as  a  basis  of  operation. 


190  INTEGRATION   OF   RATIONAL   FRACTIONS. 

The  values  of  A,  B,  and  C  may  be  obtained  from  (2),  as 
follows  : 

Making  a  =0,       we  have      3  =  —  2  A;  .-.  A  =  —  f. 

Making  x  =  1,       we  have      5  =  35;       .-.  B  =  ^. 

Making  x  =  —  2,  we  have  —  1  =  6  C ;       .\  (7  =  —  £. 

Examples. 

1.    Find  f  C*-1)^.  ^.  log  («  +  *)'«. 

J  xr  +  6x  +  8  °   (a? +  2)4 


J! 

./      XT 

s 


(il+1)t'  log[(^-l)2(x+2)«C]. 


^  +  ^-16)f-  l0g  [XKX  ~  2)KX  +  3)IC]- 

J      a? -4  SV®  +  2^ 

174.  Case  II.  When  some  of  the  simple  factors  of  the  de- 
nominator are  real  and  equal,  to  every  set  of  equal  factors,  as 
(x  —  a)n,  there  corresponds  a  series  of  n  partial  fractions  of  the 

v  A        .  B  .         ,      L 

form ;  +  •  •  •  H 

J  (x-a)n      (x-a)"-1  x-a 

Let  it  be  required  to  find   (  ; 

J  (*-l)2(*+l) 

Assume  — ■  = -H 1 •      (1) 

(x-l)2(x-j-l)      (x-1)2       x-1  x+1         v 

Clearing  (1)  of  fractions,  we  have 

1  =  A(x  +  1)  +  B(x? -  1)  +  C (x -  l)2 
=  (B+C)x2  +  (A-2C)x  +  A-B  +  C.  (2) 

Equating  the  coefficients  of  like  powers  of  x,  we  obtain 

B+C=0,   ^L-2C  =  0,    and  A-B  +  C=l.     (3) 

Whence  A=%,    C=i,   and  B  =  -\.  (4) 


CASE  II. 


191 


Substituting  these  values  in  (1),  and  integrating,  we  have 

/dx 
(x-l)Hx+l) 


~J2(x-l)2     J  4(x-l)     J  ■ 


dx 


=  loo- 


lEB+l.V 


(X- 
1 


4(a>+l) 


(5) 


\-C. 


sX—lJ         2(05—1) 

Equation  (5)  is  true  ;  for  the  values  of  A,  B,  and  C  given 
in  (4)  satisfy  (3),  and  hence  make  (2),  and  therefore  (1),  an 
identical  equation. 


Examples. 


1.    Find    f    (2*-5)*» 

J  (o;  +  3)(j»+l)a  2(a?+l)   ' 

a      C a?dx 

J  xi  +  5x2  +  8x  +  4:' 


3      r(2x*+7xi+6x+2)dx       . 
J         a:4-f-3ar3  +  2a;2  & 


7  .11,      a?+l   .   ^ 

+  log(.r+l)  +  (7. 


x  +  2 


x(x-\-l) 


x+2 


x 


175.  When  the  simple  factors  of  its  denominator  are  imagi- 
nary, a  fraction  may  then  also  be  decomposed  into  partial 
fractious  of  the  forms  in  Cases  I.  and  II.  ;  but,  since  the  inte- 
grals obtained  from  these  would  involve  the  logarithms  of 
imaginaries,  we  seek  other  forms  for  the  partial  fractions. 

Since  the  imaginary  roots  of  an  equation  always  occur  in 
conjugate  pairs,  the  imaginary  factors  of  the  denominator  will 
occur  in  pairs  whose  products  are  real  quadratic  factors  of  the 
form  (x  —  o)2  +  &2.  Hence,  when  its  simple  factors  are  im- 
aginary, the  denominator  can  be  resolved  into  real  quadratic 
factors. 

Cask  III.  WJien  some  of  the  simple  factors  of  the  denomina- 
tor are  imaginary  and  unequal,  to  every  factor  of  the  form 
(x  —  a)2+b2  there  corresponds  a  partial  fraction  of  the  form 

Ax  +  B 
(x-a)2  +  b2* 


192  INTEGRATION   OF   RATIONAL   FRACTIONS. 

x*  -f-  xr  —  2 

a  a2  A     .     B     .  Cx+D     /1N 

Assume   : -= — (1) 

(x+l)(x-l)(xr-\-2)     aj+1      x—  1      a?  +  2      yj 

Clearing  (1)  of  fractions,  we  obtain 

x2  =  A(x—1)  (x2  +  2)  +  B(x  + 1)  (x2  +  2) 
+  (Cx+D)(x2-l) 
=  (A  +  B  +  C)x3+  (B  -  A  +  D)x* 

+  (2A  +  2B-C)x  +  2B-2A-D.  (2) 

Equating  coefficients  of  like  powers  of  x,  we  have 
A  +  B+C=0,  B-A  +  D=l,  ) 
2A  +  2B-C=0,    2B-2A-D  =  0.) 


(3) 


Whence  ^  =  -|,   B  =  ±,    O=0,   JD  =  f.  (4) 

Hence       C /c[x     =  -*  f  *-+i  f*.+2  f-*L   (5) 
J  «*+«*— 2  6Js+l      6Jir-l     3Ja?+2   V  y 

=  ilog^^  +  1 V2  tan"1—  +  0. 
6     °a;  +  1^3  V2 

Equation  (5)  is  true ;  for  the  values  of  A,  B,  0,  and  D  given 
in  (4)  satisfy  (3),  and  hence  make  (2),  and  therefore  (1),  an 
identical  equation. 

Examples. 


1.    Find    f 2&L . 

J(aj+l)(a^  +  l) 

Ans.   -tur1a  +  ^log (a^  +  1)i  +  0. 
2  2    °    a?  +  l 

r       aftfa  lQg(^  +  2)c 

*   J  x4 +  3^  + 2  B(«?  +  l)» 


n      C          xrdx                          1  .  1, 

3.     I 1 — lo 

Jfj-DV  +  l)  2(a?-l)      4 


1  .  1,      a?— 2a>  +  l 


(a-l)2(y  +  l)  2(«-l)      4    °      x?+\ 


0. 


r  ardx 
J   1-x* 


CASE   IV.  !'.•:, 


-  log — ! tan1  a;  +  C. 

4     B  1  -  x      2 


h 

J  X' 


x2  dx 


+  x9  +  *  + 1 

^  log  (a  +  1)  +ilog(ar  +  1)  -  ^tan-1*  +  C. 

/"   (7a-     _  1  r  dx    _  1   r(x-2)dx . 
J  a,-3  +  1      3  J  a?  +  1      3  J  ar  -  a?  +  1  ' 

r(a?-2)c7.r=l  p(2a?—  l)dx     1  r     3cfo 
»/  ar  —  a;  +  1      2  J    ar  —  a;  + 1        2*/  ar  —  a;  + 1  ' 

.    f   c7a:         1  (1+a-)2    .      1    ,     _j2aj-l   .   „ 

.'.  I =  -log  — — L— tan  * H  C 

Ja*+1      6     Bx*-x+l      V3  V3 

p.         C     dx  ll  3?  +  «  +  l      ,       1     4.-1 2  X+l     .     r, 

7.     I -  log ! ! tan  * —  +  C . 

Jl-x*  6    ° a? -  2x  +  1      V3  V3 

176.  Case  IV.  When  some  of  the  simple  factors  of  the  denom- 
inator are  imaginary  and  equal,  to  every  set  of  equal  quadratic 
factors  of  the  form  [(x  —  a)2  -f-  b2]n  there  corresponds  a  series  of 
n  partial  fractions  of  the  form 

Ax  +  B  Cx  +  D , ,       Lx  +  M 


[(x  -  a)2  +  b2]'1      [(x  -  a)2  +  b2]""1  (x  —  a)2  +  b2 

In  any  example  under  this  case,  by  clearing  the  assumed 
equation  of  fractions,  and  equating  the  coefficients  of  like  pow- 
ers of  x,  we  should,  as  in  the  first  three  cases,  evidently  obtain 
as  many  simple  equations  as  there  are  indeterminate  quantities  ; 
and  the  values  of  A,  B,  C,  etc.,  determined  by  these  equations 
would  make  the  assumed  equation  an  identical  one. 

It  remains  to  be  proved  that  we  can  always  integrate  a  frac- 

•  «.    4  (Ax  +  B)dx 

tion  of  the  form  — -* '— — !■ — - — 

[(x-a)2  +  b2¥ 


194  INTEGRATION   OF   RATIONAL  FRACTIONS. 

Let  z  =  x  —  a  ;  then  x  =  z  +  a,  dec  =  cfe,  and 

l(x-a)2  +  b2Y~J  (z2  +  b2)n 

_  r    Azdz  r(Aa  +  B)dz 

J  (z2  +  b2)n     J     (z2  +  b2)n 

— A r(Aa  +  B)dz 

2(n-l)  (z2  +  b2)n~1     J     (z2  +  b2)n 

The  method  of  finding  the  integral  of  the  last  term  is  given 
in  §  185. 

Examples. 

1     Find   f(aP  +  X~1)dx- 
J        (ar  +  2)2 

Ana.  Uogix2  +  2)  -\ f — — 

Assnme  *f*~*  =  ^  +  J?  +  «Lt£ 

(x2-\-2)2       (x2  +  2)2       ^  +  2 

2.  I  -^ 5— ! — ^— -  log     o       \  ,  —  itan  xx  +  (7. 

J  a;3  +  a^  +  a;+l  &(a24-l)i      2 

o      /*  da;  ,      (x  —  1 )  s         1    x    _i  as    ,  /-» 

3.  I  — -•         log  f-r * tan  x h  C 

4.  fC2-^)^.  — ? 12__31og(a;+2)+C7. 


CHAPTER  XIV. 

INTEGRATION  BY  RATIONALIZATION. 

177.  It  has  been  shown  that  any  rational  differential  is  inte- 
grate ;  hence  an  irrational  differential  which  does  not  belong 
to  a  known  form  can  be  integrated,  if  we  can  rationalize  it ; 
that  is,  if  we  can  find  its  equivalent  rational  differential  in  terms 
of  a  new  variable  which  is  some  definite  function  of  the  given 
variable. 

c 

178.  A  differential  containing  no  surd  but  those  of  the  form  x5 
can  be  rationalized  by  assuming  x  =  zn,  in  ichich  n  is  the  least 
common  multiple  ofcdl  the  denominators  of  the  several  fractional 
exponents  o/x. 

For,  if  x  =  zn,  the  values  of  x,  dx,  and  each  of  the  surds,  will 
be  rational  in  terms  of  z;  hence  the  function  of  z  obtained  by 
substituting  these  values  in  the  given  function  will  be  rational. 

Examples. 

1.   Find   Cxh  ~  **  dx.  Ana.  4a*-4si  +  C. 

J      2xh  8  * 

Assume  x  =  z6 ;  then 

x*  =  z3,  x%  =  z4,  xi  =  z,  and  dx  =  6  z*dz ; 

J     2x*  J      -lz  J  v 


)dz 


=  ^-^+C  =  ^-\xi  +  C. 


2.  fj%%j  |^-i^+t^-taJ»+^rlog<Hr2a*)+C. 


196  INTEGRATION   BY   RATIONALIZATION. 

179.  A  differential  containing  no  surd  except  a  +  bx  affected 
with  fractional  exponents  can  be  rationalized  by  assuming  a  +  bx 
=  zn,  in  which  n  is  the  least  common  multiple  of  the  denominators 
of  the  several  fractional  exponents. 

For,  if  a  -f  bx  =  zn,  the  values  of  x,  dx,  and  each  of  the  surds, 
will  be  rational  in  terms  of  z. 


Examples. 

1.    Find   f -^ rv  Ans.  2tsm~1(l  +  x)h-t  C. 

J  (l  +  aj)*  +  (l  +  a>)i 

Assume  l  +  x  =  z2;  then 

(l  +  a?)i=23, 

(l+as)£  =  2,  and  dx  =  2zdz\ 

dx f2zdz  _  „  r  da 

(l  +  x)i  +  (l  +  x)l~J  z*+z~   Jz2+1 

=  2tan-12+C=2tan-1(l  +  o;)5  +  C. 


xdx  2 (2  a  +  6#)   ,  ^ 

+  tow)1  62  Va  +  &a; 


2-/r 

»/  (a 
4.     fa>(a +  »)*(&}.  ^r(4a-3a)(a  +  z)*  + C 


f      d*       .  -^logVo  +  6a?""V2  +  logc. 

•^  ^  Va  +  bx  Va       Va  +  6»  +  Va 


f   ^da?    .  3(1  + a;)* 

J  (1  +  xY*  v         y 


(1  +  aQ2     1—x 

7  2 


+  c. 


180.    A  differential  containing  no  surd  except  Va  +  bx  +  x2 
can  be  rationalized  by  assuming  Va  +  bx  +  x2  =  z  —  x. 


TRINOMIAL   SURDS.  197 


Let 

V«  +  6*  +  zr  =  z  —  x; 

then 

a  +  bx  =  z2  —  2zx, 

z2  —  a 

and 


b  +  2z 

dx  _  2(z2  +  bz  +  a)dz 
{b  +  2zf      ' 

I — r-7 — : — 5  r  -,       z-  +  bz  +  a 

Va  +  bx  +  a- 1  =  z  —  x  |  =  — ! — 

L  J  b  +  2z 


Hence,  as  the  values  of  x,  dx,  and  Va  +  bx+x*  expressed  in 
terms  of  z  are  rational,  the  given  differential  when  expressed  iu 
terms  of  z  must  be  rational. 


181.   A  differential  containing  no  surd  except  V  a  +  bx  —  x- 
can  be  rationalized  by  assuming 

Va  +  bx-x2[  =  V(x-/3)(y-x)]  =  (x-/3)z, 

in  icliich  B  and  y  are  the  roots  o/x2  —  bx  —  a  =  0. 

Let  Va  +  bx  —  x2 [=  V(#  —  B)  (y  —  a)]  =  (a;  —  /3)z ; 

then  y  —  a!  =  (#  —  /?)z2, 

22  +  l  ' 

02  +  i)2 

and  Va  +  to-arT  =  (x  -B)z]  =  ^JzDl. 


Hence,  as  the  values  of  x,  dx,  and  V«  +  bx  —  x2  expressed  in 
terms  of  z  are  rational,  the  differential  when  expressed  in  terms 
of  z  must  be  rational. 

Examples. 

1.   Find  f = An*.  log[(2a?+l+2Vl+ a? +a?)c]. 

J  Vl  +  aj  +  a,*2 


198  INTEGRATION   BY   KATIONALIZATION. 


then 


Assume    Vl+x'+^  =  2-!B; 
z°--  1 


2z  +  l 

dx  =  ni±z±})dz 
(2z  +  l)2      ' 


and  Vl  +  x  +  x!\_  =  z  —  x~]  = 

dx  _  r    2dz 

Vl  +  X  +  X2      J    2z  + 


z2  +  z  +  l 


2z  +  l 
'•  f  -=f-^=log[(2z  +  l)c] 

J  -x/1  A-r.  +  a?      J    2Z  +  1  &LV        ^     '    J 


=  log[(2  a;  +  1  +  2  Vl  +  x  +  ar^c] . 


J  y/2-X-X* 


The  roots  of  a;2  +  x  —  2  =  0  are  —  2  and  1 ; 
.-.  2  -  a;-  x*  =  (a?  +  2)  (1  -  a?). 


Assume    V2  —  x  —  x'-[_=  V(a,*  +  2)  (1—  a;)]  =  (cc  +  2)2  ; 
then  1  —  x  =  (a;  -f-  2)z2, 

1  -2z2 


dx  = 


z2  +  l' 
—  62:^2 


and  V2  —  a;  —  a? 


(^  +  1)2' 

32; 


z2  +  l 


C         dx  r—2dz      a     .  !     ,   n 

■  I  —  =  I  — =  2  cot-1z  -f  C 

J  V2  —  x  —  x2     J  z-+\ 


=  2cot-*A-Z^Y+C- 

U+2; 


3 .  r — ^ 2  cot-1  f!Lz?Y  +  a 

J  Vox-6-x2  \x~2J 


BINOMIAL   DIFFERENTIALS.  199 

182.  Binomial  Differentials.   Differentials  of  the  form 

af  (a  +  bxn)pdx, 

in  which  m,  n,  and_p  represent  any  numbers,  are  called  binomial 
differentials.  When  p  is  a  whole  number,  the  binomial  factor 
can  be  expanded,  and  the  differential  exactly  integrated  by  pre- 
vious methods.     In  what  follows,  p  is  regarded  as  fractional ; 

and,  in  the  next  section,  we  will  represent  it  by  -,  r  and  s  being 
whole  numbers. 

r 

183.  Conditions  of  Rationalization  of  xm(a  +  bxn)*dx. 
I.    Assume  a  +  bxn  =  z" ; 

then  (a  +  bxn)^  =  zr ;  (1) 

._£=-£   *"  =  ^f;  (2) 


and  d»=— 2-^^-^V      dz.  (3) 

Multiplying  (1),  (2),  and  (3)  together,  we  obtain 

ro+l 

xm(a  +  bxn)!Tdx  =  —  zr+>-1(z^^\n       dz.  (4) 

bn  \     b     J 

The  second  member  of  (4)  is  rational,  and  therefore  integra- 

ble,  when  — —  is  a  whole  number  or  zero. 
n 

II.    Assume  a  +  bxn  =  z"xn; 
then  xn  =  a(z'  —  ft)"1;  (1) 

1  lmm 

x  =  a~>(z'  —  b)    »,  X™  =  an (z"  —  b)~  ;  (2) 

and  dx  =  --<^z'-1(z'-byi-1dz.  (3) 

n 

Multiplying  (1)  by  6,  and  adding  a,  we  obtain 

a  +  bxn  =  -^-  ; 
z'  —  b 


200  INTEGRATION  BY  RATIONALIZATION. 

/.  (a  +  bxny  =  aJ(z°  —  b)~^zr.  (4) 

Multiplying  (2),  (3),  and  (4)  together,  we  have 

of  {a  +  bxn)Tclx 


»+l  ,  r 


■-m-CS^O 


?i 
The  second  member  of  (5)  is  rational  and  integrable,  when 

m  ~\~  1        9* 

— _ — 1_  _  is  a  whole  number  or  zero. 
n         s 

r 

Hence,  xm(a  +  bxn)^dx  can  be  integrated  by  rationalization. 

*???      1      1    * 

I.  When  — -L_  is  a  whole  number  or  zero,  by  assuming 
a  +  bxn  =  z\ 

II.  When  — — — (--  is  a  whole  number  or  zero,  by  assuming 

n         s 

a  +  bxn  =  z'xn. 

Examples. 

1.   Find  Cx^ia  +  btfyhdx.        Am.  h:x?  ~?a  (a  +  bx*)h  +  C. 
*/  35" 

Here  — —  is  a  whole  number,  and  s  =  2  ;  hence  we  assume 
n 

a  +  bx2  =  z2; 
.'.{a  +  bx£)-h  =  z-1;  (1) 

x=(t^]\    *-f*=SYi  (2) 


?7l  -4-  1  772  -4-  1         r 

*  When 1- 1  is  a  negative  integer,  or 1 \- 1  is  a  positive 

integer,  the  exponent  of  za— b  being  negative,  the  given  differential  will  be 
reduced  to  a  rational  fraction  whose  integral  may  be  obtained  by  the 
method  of  Chapter  XIII.  But,  as  this  method  usually  gives  a  complicated 
result,  it  is  generally  expedient  in  such  cases  to  integrate  by  using  the 
formulas  of  reduction  given  in  §  185. 


EXAMPLES.  201 

and  dx=-(— )dz.  (3) 

b  \z-  —  a  J 

Multiplying  (1),  (2),  and  (3)  together,  we  obtain 
fa?(a  +  bx*)-hdx  =  \  C(z- -  a)dz 

=  ^{z^-Zaz)  +  C=h^^{a  +  b^  +  C. 


4-3ar>         c 
i  9(2 -3ar')* 


o      C     a?dx 
J  (2-3.T2) 

3.  fx-*Q  +  a?)-iax.  (2^-l)(l  +  ^  +  a 
J  3  or3 

???/     I      1  7' 

Here  — — — 1_  _   is  a  whole  number,  and  s  =  2  ;    hence  we 
?i  s 

assume 

l+a^  =  z2#2;  ,-.  xr  =  (z2  —  1)~\ 

0.(*-l)-ft,  ar<  =  (z2-l)2,  (1) 

dx  =  -(z2-\)-izdz;  (2) 

and  (i  +  ar!)-l=/'l  +  -Ti-y"=^  x(22-l)-i  (3) 

Multiplying  (1),  (2),  and  (3)  together,  we  have 
J  x4(l  +  ar9)5         J  v  J 

3  3a;3 

4.  I ^ — — •  -log : f-logc. 

J  X(cr  +  X- )  *  a  Vrr  -+-  .of  4-  ™ 


/dec                                        1,                a;  .  , 

-71 ^T"  -log-— = hlogc. 


5. 

;(a2-ar)*  a       y  a*  -  or  + 


C     ad.r 
J  (l+ar)i 


(l+a?)» 


+  C. 


202  INTEGRATION  BY  RATIONALIZATION. 

Miscellaneous  Examples. 

1.  Find  (•('-*>,"'■ 

J       1  —  x* 
Ans.  6[i.TB  +  fa*  - 1 a?i  +  \a&  —  fa*  +  a*  —  log(l  -f-  a;*)]  +  (7. 


2.  I losj- l -Hogc. 

3.  f *? .  2tan-1(l  +  aj)l+C. 

J  (2+oj)(l  +  a;)l  V   T    7 

4.  I  (ft  +  te)8o;dr.  -i — '— — '—I — )  +  C. 

_       C     ®dx  3  /     i  t,  \  a  /ft  +  bx     «\  ,   sy 

»/a;(&a;  — ft)s  -^  \     a     J 


clx _  2  /2-aV 

(l  +  aj)V2-|-a5-aja  8\*  +  l 


8.  f cfa  lQg/  a?+VT+^+g_c\ 

J  (1  +  aOVl  +  aj  +  aj*  \2  +x  +  Vl  +  a:  +  ar2  J 

9.  Ca?(a  +  ba?)%da;.  (ft  +  to2) sf5  6^  ~  2 ")  +  C 

10     f     ^  «(**+•*)  ,  c 

"   J  (1  +  aJ8)!  3(l  +  ar>)5 

ii    f  ^^  x3       i  (7 

J  (ft4-^)i#  8a(o  +  6a^)i 

12.     f * q  +  26a*     +  C. 

J  ar(«  -f  for)l  «2a;(a  +  fov2)  2 


CHAPTER   XV. 

INTEGRATION  BY  PARTS  AND  BY  SERIES. 

184.   If  u  and  v  be  any  functions  of  a;,  we  have 

d(uv)  =  udv  -f-  vdu. 
Integrating  and  transposing,  we  have 

I  udv  =  uv  —  J  vdu.  (1) 

Equation  (1)  is  the  formula  for  integration  by  parts.  It 
reduces  the  integration  of  udv  to  that  of  vdu  ;  and,  by  its  appli- 
cation, man}'  useful  formulas  of  reduction  are  obtained. 

Examples. 

1.  Find   j  x  log  xdx.  Ans.  fax*  log  a  —  \x* -\-  C. 
Assume  u  =  log  a;,  and  dv  =  xdx ; 

then  du  =  — ,  and  v  =  — 

x  2 

Substituting  these  values  in  formula  (1).  wo  have 

j  x  log  xdx  =  -%x?  log  x  —  I  ^iceZa; 

=  ^-x2  logo;  —  ^ar  -J-  C. 

In  each  example,  the  values  of  u  and  dv  must  be  so  assumed 
that  vdu  is  a  known  form,  or  nearer  one  than  udv. 

2.  |  log  xdx.  #(loga;  —  1)  +  G. 

3.  Cxe^dx.  erf-  -  -VW  C. 
J  \a      a- J 

Assume  dv  =  e^dx. 


204  INTEGRATION   BY   PARTS. 

4.  I  xnlog:xdx.  (los^x \  +  C. 

J         &  n  + 1\    °         n+\) 

5.  I  s\n~lxdx.  X  &\K~l  X  ■{- (1  —  X2)h+  C. 

6.  I  tsxrrlxdx.  xt&n^x  —  log ( 1 -f- ic2) 3 -f  C. 

7.  Cx\a  -  ar^dx.  -  \a?(a  -  b2)!  - ^%(a -  x2) f  +  C. 
Assume  it  =  x2. 

8.  |#cosa;da;.  a?  sin  x  +  cos  #  -+-  C. 

185.  Formulas  of  Reduction.  We  will  next  apply  the  formida 
for  integration  by  parts  to  the  binomial  differential, 

xm(x  +  bxn)pdx* 

in  which  p  is  any  fraction,  but  m  and  n  are  whole  numbers,  and 
n  is  positive. 

I.    Let  dv  =  (a  +  bx^x^dx,  and  u  =  xm~n+l ; 

then  -y  =  -*> — — ' — ,  and  du  =  (m  —  n  + 1 )  xm~n  dx. 

nb{p+l)  K  t  ) 

Hence,  by  the  formula,  we  have 

C  m/     ,;   »\»  7        xm-n+1(a  +  bxn)p+1 
I  xm(a  +  bxn)pdx  = i — ! L — 

J       V  J  nb{p  +  \) 

-  m~n+1  Cxm-n(a  +  bxnY+ldx.  (1) 

nb(p+l)J  K  '  K  J 

*  That  any  binomial  differential  can  be  reduced  to  one  of  this  form  is 
evident.  For,  let  the  given  differential  be  (x~*  —  x*)*  x~*  dx ;  then,  by  mul- 
tiplying the  first  factor  and  dividing  the  second  by  (x*)*,  we  obtain 
(1— xB)3a;— E  dx.  Putting  x  =  z6,  we  have  6  (1  —  zb)*  dz,  which  is  of  the  form 
required. 

The  formulas  of  reduction  are  true  for  fractional  and  negative  values 
of  m  and  n,  but  they  are  not  generally  useful  in  leading  to  known  forms, 
unless  m  and  n  are  whole  numbers,  and  n  positive. 


FORMULAS   OF    REDUCTION.  205 

Now  Cafl-n(a  +  bxn)p+1dx  =  C xm~n(a  +  bxn)»(a  +  bxn)dx 

=  a  Cxm~n(a  -f  bxn)pdx  +  b  Car  (a  +  6a;n)pcfa. 

Substituting  this  value  in  (1),  we  have 

Car  (a  +  bx"ydx  =  *"-"+1(«  +  6*r+1 

_  (m-n+l)a  ^  .  „(a  +  ^^ 

-  ??l~w  +  1  (V(«  +  6a;-)  "cte. 

Transposing  the  last  term  to  the  first  member,  and  solving 
for  I  xm(a  -+-  bxn)pdx,  we  obtain 

I  of  (a  +  bxn)vdx  = i — ! '- — 

J  b(iip+m+\) 

-  «(™-"+D    (>»(a  +  j^ifa  (A) 
b(np  +  m  +  l)J         v            y  v    ; 

By  formula  (A)  the  integration  of  of1  (a  +  bxn)pdx  is  made  to 
depend  upon  that  of  another  differential  of  the  same  form,  in 
which  m  is  diminished  by  n.  By  a  repetition  of  its  use,  m  may 
be  diminished  by  any  multiple  of  n. 

Formula   (A)   evidently  fails  when  np  +  m  + 1  =  0  ;    but  in 

1Yh  *4"  1 

that  case (-  P  =  0  >  hence  the  method  of   integration  by 

rationalization   (§  183)   is   applicable,  and  the  formula  is  not 
needed. 

II.    It  is  evident  that 

Car  (a  +  bxn)pdx  =  Cxm(a  +  bxn)p-1(a  +  6a,-")cZa; 

=  ai  xm(a  +  bary^dso 

+  6  f a;m+"(a  +  6xn)'-1  cte.  (2) 


206  INTEGRATION   BY   PARTS. 

Applying  formula  (A)  to  the  last  term  of  (2) ,  by  substituting 
in  the  formula  m  +  n  for  m,  and  j;>  —1  for  jp,  we  obtain 

b  Cxm+n (a  +  bx*)*-1  clx  =  xm+1(a  +  bxnY 
J  np  +  m  -f- 1 

q(m  +  l)    (V(a+6af»y-adaj. 

np  +  m  +  1 J 

Substituting  this  in  (2) ,  and  uniting  similar  terms,  we  have 


/' 


'of*  (a  +  bxnYdx  =  a;m+1(a  +  &>?)* 
tip  +  m  + 1 


anp 


-  Cxm(a-\-bxny~1dx.  (B) 


wp  +  m-j- 

Each  application  of  formula  (B)  diminishes  p,  the  exponent 
of  the  binomial,  by  unity.  It  fails  in  the  same  case  that  (A) 
does. 

III.  When  m  and  p  are  negative,  we  need  formulas  to  increase 
rather  than  to  diminish  them.  To  obtain  these,  we  reverse  for- 
mulas (A)  and  (B). 

Solving  (A)  for  I  xm~~n(a  +  bxn)pdx,  and  substituting  m  +  n 
for  m,  we  obtain 

C  »/     n  n\v  7        xm+1(a+bxn)p+1 

I  of1  (a  -f  bxnY  ax  = ^ — ! - — 

J      v  '  a(m+l) 

_  6(nj)+w+m+l)  r  m+;i(a  +  &<)^  (C) 

a(ra  +  l)         J 

Formula  (C)  enables  us  to  increase  m  by  n  at  each  applica- 
tion. It  fails  when  m  +  1  =  0  ;  but  in  that  case  the  differential 
can  be  rationalized  (§  183). 

IV.  Solving  (B)  for  J  xm(a-\-bxn)p~ldx,  andsubstitutingp+1 
for  p,  we  obtain 

Cxm(a  +  bxnV  dx  =  -  a""+1(a  +  k*")p+1 
J       v  y  an(p  +  l) 

np  +  w  +  m+1  p     (        hr^dto.  (D) 


EXAMPLES.  207 

Formula  (D)  enables  us  to  increase  p  by  unity  at  each  appli- 
cation. 

The  mode  of  applying  formulas  (A),  (B),  (C),  and  (D)  will 
be  illustrated  by  a  few  examples. 


Examples. 


1.    Find  f    fd0Do    • 


(a2 -a8)! 

Here  m  =  4,  n  =  2,  p  =  —  •£,  a  =  a2,  and  b  =  —  1 .  Hence,  by 
applying  formula  (A)  twice  in  succession,  this  iutegral  will  evi- 
dently be  made  to  depend  upon 

dx 


f 


V«2  —  x2 

Substituting  these  values  of  m,  n,  etc.,  in  formula  (A),  we 
obtain 

J  x*(a2  —  x2)~*dx  =  —  %x?(a2  —  x2)^ 
-f  f  a2  J  ar(a2  —  aF)~ldM.  (1) 

In  like  manner,  we  obtain 

I  ^(a2  —  x2) ~i  dx  =  —  -j #(a2  —  x2) I 
+ia2f(a9-«!,)-idaj.  (2) 

Now,         f(a2-ar)-kto=  f_-J|_  =  sin"1-  +  C.         (3) 
J  J  ^/tf-x2  « 

From  (1),  (2),  and  (3),  we  obtain 

C   xUJx     =  -  (±a?  +%a2x)  Va2  -  x2 
J  V«2  —  x2 

8  a 

2      f^tf^.  ^   _^V^^?  +  ^sin >*  +  <?. 

J  Va2-^  2  2  a 


208  INTEGRATION  BY  PARTS. 

3.  f    ^dx    ■  -\(a?+2a2){a?-x>)*+C. 
J  Va2  —  x2 

/*cm  (Tic 
—      '       is  made  to  depend  on  what  known 
Va2  —  ar 
f orm  when  m  is  positive  and  even  ?     On  what  known  form  when 
m  is  positive  and  odd? 

4.  C    ^clx     •  -VaM^-*a2log(x-+Va2+V)  +  (7. 

J  Vtf  +  tf         2 
5  _    r  a*dx    t  i  ^  _  2  ^  ^/^tj— i  +  a 

•^  Va2  +  cc2 

C*  x™  dx 
By  formula  (A) ,  I     '  is  made  to  depend  on  what  known 

form  when  m  is  positive  and  odd?     On  what  known  form  when 
m  is  positive  and  even? 

-      f*    a;6  da; 

^  Va2  —  x2 

fx5   .  5a2x*  ,  5-3a4x\   /-» ,  ,    5 -3a6  .  _rx  ,   r, 

— Va-  —  ar  H sin    -  +  C 

\6        6-4        6-4.2y  6-4.2  a 

8.  C{a2-x2)Ux.  bx(a*-x*)h  +  $a*8m~1-+C. 

*J  CI 

Appl}-  formula  (B)  once. 

9.  Cx2(l—xi)idx.  ix(2x2-l)(l-x2)h  +  %sin-1x  +  C. 

10.     fx*(l-x*)idx. 

i(i^-iV^-l^)(l-^)'+i16sm-1a;+(7. 

11       C        dx  Va2  — ar      p 

*   Jar* (a2 -a,-2)*'  a2a;       "^ 


EXAMPLES.  209 


12.     f * Vcr- or      J_,      X 

J  a?  (cr  —  ar) I  2  crar        2  a*       yV  _  ^  +  a 

For    f      c^        see  §  183,  Ex.  5. 

Apply  formula  (B) . 

14.  C(c?+a?)kdx.      %xVaF^+$anog(x+VaY+tf)+C. 

15.  f— ^ ? +  — ,  tan-^  +  C. 

J  (a2  +  a?) 2  2  a2 (a8  +  aj2)      2  a3  a 

Apply  formula  (D) . 

16.  ^(l-^ida;.      £a>(l— a*)i-|-£a>(l— a?)*+f  aur^BH-C. 

17.  f      xdx       ■  _  (2  o.r  -  a?)  I  +  o  vers"1-  +  C. 
•^  V2  ax  —  x2  a 

r*      xdx  f 

Write  I      ,  ■=  in  the  form  I  xi(2a  —  x)~*dx,  and  applv 

J  -\/2ax  —  x2  J 

formula  (A)  once. 

18.  I — ■■  '- — v2aa;  —  x--\-  fevers1-  +  C 

J  -\Z2ax-x2  2  a 

u.  f  *"   . 

*^  V2  aa  —  xr 


,-jft. 

o  a 


_  2x2  +  5a(x  +  Sa)  V^x~^  +  |a3vers-i*  +  c< 
20.     f      * 


1 8 


? I I* i_  JL  tin-1-  -I-  C 

4a8(as  +  a2)2 T  8«4(«2  +  a8)      8a5  '       a 

21.     f    ^    .  *(3-ar°)  , 

J  (i-^)a  2(1-^)4    *sm   X+C- 


210  INTEGRATION   BY   PARTS. 

186.  To  integrate  the  logarithmic  differential  <£(x)  (7o^x)"dx, 
in  which  </>(x)  is  an  algebraic  function,  and  n  is  a  jiositive  whole 
number. 

Let  a^  (log  a;)2  da;  be  the  function. 

Assume   dv  =  x2dx,  and  w  =  (loga;)2; 

then  du=2logx — ,  and  v  =  ^x?. 

x 

Hence,  by  the  formula  for  integration  by  parts,  we  have 

J  a^(loga;)2da;  =  ^ar3(loga;)2  —  f  I  a;2  log  a;  rite.     ■    (1) 

Applying  the  formula  to  the  last  term  of  (1),  we  have 

J  ar  loga;rite  =  ^  a,*3  log  a;  —  I  ^a?dx.  (2) 

From  (1)  and  (2),  we  obtain 

Ja^(loga;)2da;=^[(loga;)2-|loga;  +  |]+(7. 

Hence,  to  integrate  <f>(x)  (log x) ndx,  we  assume  dv  =  <f>(x)dx, 
and,  by  successive  applications  of  the  formula  for  integration 
by  parts,  reduce  the  exponent  of  log  a;  to  zero.  In  this  way 
the  integration  of  the  logarithmic  differential  is  reduced  to  the 
integration  of  algebraic  differentials. 

Examples. 

1.  Find   Cxz{\ogx)-dx.      Ans.  ia;4[(loga;)2-^loga;+|]  +  C 

0      /*  log  a;  rite  x    ,  ,     /1  .     N  .   „ 

2.  I  — ^ — - •  ■  log  a;  —  log  ( 1  +  x)  +  C. 

J  (1+a;)2  1+a;    °  5V  ; 


3.     |  a;"  (log xfd 


x. 


n  +  1 


9  2 

(loga;)2 ^-loga;+ 


n+1     °        (n  +  l)2_ 


+  C. 


(lo^ar)2  2 

— <*»;  -^^[(log^)2  +  |loga;  +  |]+C. 


EXPONENTIAL  DIFFERENTIALS.  211 

187.  To  integrate  the  exponential  differential,  xnamxdx,  when 
n  is  a  x>ositive  integer. 

Assume   civ  =  a^clx,  and  u  =  xn ; 

a""" 

then  cht  =  ?iajn_1c?ic,    and   v  = 

m  log  a 

Hence,  by  the  formula  for  integration  by  parts,  we  have 

/xnamxdx=  af>C(mJ —  Car-1ar'dx. 
mloga      vilogaj 

By  successive  applications  of  this  formula,  the  exponent  of  x 
is  reduced  to  zero,  and  the  proposed  integral  is  made  to  depend 
upon  the  known  form 


/■ 


cfdx. 

Examples. 


1.  Find   (xre^dx. 

Assume  dv  =  efdx,  and  u  =  x2 ; 

then  (  are<"cte  =  -e^z? (  xe"dx. 

J  a  aj 

Again,      j  xe^dx  =  -e"x J  e^dx. 

Hence       j  x*eaxdx  =  —  [x2 h  —  )  +  C 

J  a  \  a      ay 

o      r5K]  e"/*      3  o  .  3-2        3-2-l\  ,   „ 

2.  I  7feaxdx.  — (or xr -\ rx — )+ C. 

J  a  \         a  a-  a3    / 

q      C*  xn  o1   Tus      8. a*  .3.2-3        3-2-1  "1  ,  ^ 

3.  I  ara'dx.  x? 1 +  C. 

J  l°ga[_         loga      (loga)-      (loga)3J 

4.  Write  out  the  integral   of  exx4dx,  according  to  the  law 
of  the  integrals  in  Examples  2  and  3. 

188.    To  integrate  the  trigonometric  differential  smmx  cosnx<I\\ 
Let  sin  x  =  z ; 

then  sinm.r  =  zm,  cosncc  =  (1  —  z2)  2", 


212  INTEGRATION  BY   PARTS, 

and  dx  =  (l  —  z~)-*dz. 

/f*  n— 1 

sinma; cos" x dx  =  I  2^(1  —  z2)~2~dz.  (1) 

Or,  letting  cos  x  =  z,  we  obtain 

J  sinm x  cos" as dx  =  (  —zn(l  —  z2)~dz.  ( 2 ) 

Hence,  whenever  the  binomial  differential  in  (1)  or  (2)  can 
be  integrated,  the  given  differential  can  be. 

This  method  of  integrating  sinm£  cos"  xdx  is  used  in  those 
cases  to  which  the  shorter  methods  of  §  63  are  not  applicable. 

Examples. 

1.  Find   J  sin6 xdx. 

Put  sin  x  =  z; 

then  dx  =  (1  —  z2)~*dz, 

and  I  s'mGxdx=  I  z6(l  —  z2)~*dz 

+  -^-sin-^  +  C  §  185,  Ex.  7. 

D  •  4  •  I 

cos a/  .  s     ,   *   •  3     ,5-3.      \ 
= sin3a;  +  f  suraH sm# 

6  v  4-2     y 

6-4-2 

2.  I  sin4 x dx.  Ans.    —  icosa;  (sin3cc  +  f  sina)  4-  f  a;  +  C. 

3.  |  cos*  a;  da?.  ^sina,*(cos3a;  +  f  cosa:)-t-f x+  C. 

4.  I  sin2  x  cos2  a;  dx.  -|- sin3  a;  cos  a;  —  |-sina;cosa/'-|--g-#4-  C, 

or  i(a;  — isin4a;)-f  C 
Let  since  =  z,  and,  for    j  z2  (1  —  22)5C?z,  see  §185,  Ex.  9. 


TRIGONOMETRIC   DIFFERENTIALS.  213 

The  second  form  of  the  integral  is  obtained  from  the  first  h}' 
use  of  the  relations,  2  sin x cos  x = sin  2  x,  and  2siu2#=l  —  cos  2a;. 
By  a  similar  transformation,  any  differential  or  integral  expressed 
in  powers  of  sin  x  and  cos  x  may  be  found  in  terms  of  the  sines 
and  cosines  of  multiples  of  x. 


4    ,  sin  x  /cos3  x     cos5 a;  .  cosaA  .    x    ,  „ 

sin-  x  cos4 x ax.       [ \-\ \-  C. 

2    \   12  3  8    J     16 


r dx 

J  sin.rcc 


-f-  log  tan  x  -f  C. 


cos3  a;  2  cos- a; 

dx  cosx     ,    ,  ,      ,      a;  .   n 

— — •  — —  +  £logtau-  +  C. 

sin.r  2  sin- a;                     2 


COS   Oj  Cl*C  i^i  i    i  &    i     s~i 

4-  cos-3  a;  +  cos  x  -f  log  tan  -  +  C  • 

sin  a)  2 

189.  To  integrate  x"sm(ax)dx,  awcZ  xncos(ax)dx. 
Assume  «  =  a,*",   and  apply  the  formula  for  integration  by 

parts.  Each  application  of  the  formula  will  evidently  diminish 
n  by  unity ;  hence,  when  n  is  a  positive  integer,  the  integral 
can  be  made  to  depend  on  the  known  form 

|  cos  (ax)  dx  or    j  sin(aa;)da?. 

190.  To  integrate  eaxsm"xdx,  and  eaxcosnxclx. 
Put  dv  =  e^dx,  and  u  =  sinna; ; 

then  v  =  -eax, 

a 

and  du  =  n  sin"-1  a;  cos  x  dx. 


f 


eaxsmnxdx  =  -e<"sin"a; 


--S 

aJ 


a 
e<™s[n"-iXcosxdx.  (1) 


Again,  put  dv  =  e^dx,  and  u  =  sin"  * x cos x ; 

then  v  =  -  e"*, 

a 


214  INTEGRATION   BY   PARTS. 

and  du  =  (n  —  1)  sinn_2a;  cos2xdx  —  smnxdx 

=  (n  —  1)  sinn~2a;cZa;  —  n  sin"  a; da;. 

[Since  cos2a;  =  1  —  siirx'.] 


'•/■ 


eox  sin  n-i  x  cos  xdxz=-eax  sin"  l  x  cos  a; 


—  |  eax  sin"  2a,-c7a;-|--  |  e"*  sinnxdx. 


Substituting  this  result  in  (1),  and  solving  for  J  e"  sinna,*c?.r, 
we  obtain, 

eax  gmn  x  CJX  _ V / 

iv  -f-  cv 
n(n-l)  feaxSin"-2a;cte.  (2) 

By  repeating  this  process  or  the  application  of  this  formula, 
n  is  reduced  to  zero  or  unity ;  and  the  integral  is  made  to  de- 
pend upon  the  known  form  I  e^dx  or  the  form  j  e^sinxdx. 
The  value  of  the  latter  form  is  obtained  directly  from  (2)  by 
making  n  =  1 . 

/• 


In  like  manner   (  eax  cosxdx  can  be  obtained. 


Examples. 

1.  Find   I  as2  cos  a;  da?. 

Here  u  =  x2,  dv  =  cosxdx, 

v  =  sin  a;,  and  du  =  2xdx. 

.•.  j  x2  cosxdx  =  a;2  sina;  —  2  j  x  sinxdx 

=  cc2  sin  a; +  2  a;  cos  a;  —  2  I  cosxdx 

=  a^  sina;  +  2a;  cosa;  —  2  sin  a;-}-  C. 

2.  I  a,-3  sin  x  dx.       —  x3  cos  a; +3  a^sinaj-f  6  x  cos  a;— 6  sin  x  +C. 


TRIGONOMETRIC   DIFFERENTIALS.  215 

/eax 
e"  s'mxdx.  — (a  sin#  —  cosx)  +  C. 

a-  +  1  v  y 

/'  ex 

exs'msxdx.       — (sin3x4-3cos3x4-3sinx  — Gcosx)  +  C. 

dx 


191.    To  integrate 


r    dx         r 

J  a+b  cosx     J 


a  -\-b  cosx 

dx 


+  6  COSX       J       (         oX  ,      .    oX\    ,    jf        oX         .    oX\ 

a  I  cos- — hsiir-  )-\-b[  cos- sin-- 


-s 


dx 


(a  +  b)  cos2!  +  (a  -  b)  siir| 


=/ 


sec2-dx 


+  &)  +  (a-&)tan2| 


-/■ 


tan  | 


(a  4-  &)  4-  (a  —  &)  tan8- 

2 


•which  is  readily  reduced  to  the  known  form, 

/-— — -  or    (  — — — -,  according  as  a  >  or  <  b. 
r  +  or       J  cr  —  ar 


+ 

f- 

+  &  sin  x 


In  like  manner   { can  be  found. 

J  a 


192.    To  integrate  the  anti-trigonometric  differentials, 

f  (x)  sm^xdx,  f(x)cos-1xdx,  f  (x)  £an_1xdx,  etc., 

in  icliicli  f  (x)  is  an  algebraic  function. 

Assume  dv  =/(x)  dx,  and  apply  the  formula  for  integrating 
by  parts.  One  application  of  the  formula  will  evidently  make 
the  integral  depend  on  an  algebraic  form. 


216  INTEGRATION   BY   SERIES. 


1.   Find  f-S^-tan-1*. 

J    1  +  SB8 


Examples. 

ft 

+  ■ 


Here  cm?  = =  dx -,  u=  tan  1x, 

\+x>  1+x2 

v  =  x  —  tan_1a;,  and  du  = 


1+x2 

—  tan_1#  =  a  tan-1  ct'  —  (tan-1  a;)2 

1+x2  v  ' 

C  xdx        rta,n-1xdx 
J  1+x*     J      1+x2 
=  a;  tan-1  a;  —  (tan-1  a;)2 

-  ilog  (1  +  x2)  +  i  (tan-^)2+  C 


=  tan-1  a;  (x  —  \  tan-1  a;)  —  log  Vl  +X2  +  C. 

2.  I  xcos^xdx.        £  jc2  cos-1  a;  —  £#(1  —  x2)?  -f-isin_1a;  +  C. 

3.  I  arsin-1a;daj.  ^a^sin-1#  +  ^(x*  +  2)  Vl  —  x2+C. 

193.  Integration  by  Series.  When  we  cannot,  by  any  of  the 
preceding  methods,  integrate  a  given  differential  exactly  ;  or, 
when  the  integral  obtained  by  them  is  of  a  complicated  form, 
we  can  develop  the  given  differential  in  a  series,  and  integrate 
its  terms  separately.  Moreover,  integration  by  series  furnishes 
a  simple  method  of  developing  a  function,  when  we  know  the 
development  of  its  derivative.  For  examples  of  this  method  of 
developing  functions,  see  §§107,  108. 

Examples. 
1.   Find    fa^(l  — oj^ldo?. 

(1-  ar)s  =  1  -  \x?  - \x'  —  -j^a-6 ; 


EXAMPLES.  217 


'.  fa*(l— a?)ida?=  Cxh(l-^xi-ix*-1\xa ) 


dx 


which  is  the  required  integral  for  a;  <  1  and  >  —  1. 


2. 


JV(1  -  Oidn.         ;V.r  -  TV^ -  ^af  -  rh^  -  ».  +  C 


3.    Prove  that  log  (a  +  x)  =  log  a  +  -  -  —  +  — —  + 

a      2d1      da3      4  a* 

dx 

by  first  integrating  directly,  and  then  by  series. 

a-\-x 

dx 


4.    Develop  log  (a;  +  Vl  +  ar')  by  integrating 


Vl+aj" 


Ans.  log(aj+Vl+iB2)  =  aj--  ^  +  —  --^^  — +■ 
oV  7  2   3       2-4  5      2-4-6   7 


5.    Prove  that 


/i 


dx                1  x5  .   1  -  3  a*     1-3-5  a;13  ,        .  n 
-  =x h----f-0. 

k  O    £  ')  .  A     Q  O  _  ,l   .  ft  1  Q 


(1+a:4)*  25       2-49      2-4-6  13 

194.  While  we  can  differentiate  any  given  integral,  we  can 
exactly  integrate  but  a  small  number  of  differentials.  One 
great  reason  for  this  seeming  difference  in  the  perfection  of  the 
two  branches  of  the  Calculus  is  that  the  integral  is  often  a 
higher  or  more  complex  function  than  its  differential.  Thus, 
the  differentials  of  log  a;,  sin-1fls,  tan-1  a?,  etc.,  are  algebraic  func- 
tions. Moreover,  it  is  evident  that  certain  forms  of  differentials 
do  not  arise  from  the  differentiation  of  any  known  functions. 
Hence,  to  obtain  the  exact  integrals  of  many  differentials,  new 
and  higher  functions  must  be  invented  and  studied.  The 
integrals  of  these  differentials,  as  obtained  by  series,  are  the 
developments  of  these,  as  yet,  unknown  functions. 


CHAPTER  XVI. 

LENGTHS  AND  AREAS  OF  PLANE  CURTES,  AREAS  OF  SUR- 
FACES OF  REVOLUTION,  VOLUMES  OF  SOLIDS. 

195.  Examples  in  Rectification  of  Plane  Curves.  For  formulas, 
see  §  65. 

1.  Find  the  length  of  the  parabola  y-  =  2px. 

Here  s  -/ 0 +S)1*-/ (* + J)'dy= jjv+^i* ; 

...  ,  =  y^E±t+P\og(y+  Vi?+F)+  C.  i  185,  Ex.  14. 
2,p  l 

If  s  be  measured  from  the  origin,  s  =  0  when  y  =  0,  and 

£=  — £j>logj>. 

,..-XV?+?+|k,t!^±P. 

2.  Rectify  the  circle  a^-f  y2  =  y-2. 

We  use  L  to  represent  the  entire  length  of  any  closed  curve. 

Here       L  =  4  f  Yl  +  ^Ycfce  =  4  r  f     r?,T      =  2  ttt. 
Jo  ^       dary  Jo  Vr2  _  a^ 

For  the  value  of  7r,  see  §  107. 

3.  Rectify  the  ellipse  y2  =  (1  —  e2)  (a2  —  or) . 
Here      ^  =  -(1-  r)  *  =  _^V|H?; 

(fas       /       e2ar        eV  3e6a^ 


Jo  Va- 


2a      2-4a3      2- 4 -6a5 


RECTIFICATION    OF   PLANE   CURVES. 


210 


Jpa    dx  2e2  ra    afdx      _  gj    /~»  aslc/aj 

'  aaJo  -y/cr  —  or 


3e6 
2-6( 


For  the  indefinite  integrals  of  the  last  three  terms,  see  §  18"), 
Examples  1,  2,  and  7.  Finding  the  definite  integrals  between 
the  given  limits,  and  adding  the  results,  we  have 

3e4         32.5eB 


L  =  -277u[\-- 


22. 42      22.4--G- 


4.  Rectify  the  hypocycloid  osl-\-yl  =  al. 

5.  Rectify  the  tractrix. 

The  characteristic  pro- 
perty of  the  tractrix  is  that 
the  length  of  its  tangent  pt 
is  constant.  Denote  this 
constant  length  by  a  ;  and 
let  o  be  the  origin,  oa  being 
the  tangent  at  A ;  then,  if 
PM  =  ds,  —  PN=cty,  NM  =  dx-, 

ET 


Ans.  6  a. 


■y/aF—tf. 

TT             ds           PM 
Hence    —  = = 


Fig.  54. 


dy 


PN 


also,  -=/  =  — 

dx 


Vcr  -  y- 


(1) 


.\s  =  —  a  I  —  =  a  logf  -  ),  if  s  be  measured  from  a. 

J  y  \yJ 

In  this  example  we  have  found  the  length  of  a  curve  without 
knowing  its  equation.  For  the  length  of  the  catenary  obtained 
in  a  similar  way,  see  §  7G,  Ex.  8. 


196.    To  rectify  a  curve  given  by  its  polar  equation. 

Since       ds2  =  pW  +  dP2,  §  136  (3) . 


220  LENGTHS   AND   AREAS   OF   PLANE   CURVES. 

Examples. 

1.  Rectify  the  spiral  of  Archimedes,  p  —  ad. 

Here        —  =  -;    .'.«  =  -  (  (a2 -\- p2)^dp. 
dp      a  a  J 

Hence,  if  s  be  measured  from  the  pole,  we  have 

s  =  p(g'  +  P»)*     a.     p  +  V^+7      §185,  Ex.14. 
2  a  2     °  a 

This  is  equal  to  the  arc  of  the  parabola,  y2  =  2  ax,  intercepted 
between  the  vertex  and  the  point  whose  ordinate  equals  p 
(§195,  Ex.  1). 

2.  Rectify  the  logarithmic  spiral  p  =  ae. 

Here        s=  1  ( 1  +  m2)  h  dp  =  ( 1  -f  ?>i2)  £p, 

in  which  ??i  is  the  modulus  of  the  system  of  logarithms  whose 
base  is  o,  and  s  is  measui-ed  from  the  pole. 

3.  Construct  and  rectify  the  cardioid  p  =  a(l+  cos0) . 

L  =  2  f[a2(l+  cos0)2  +  a2  sin20]id0 

Q   cn  ode    Q 

=  8  a  I  cos =  8  a  ; 

Jo        2  2 

since  2(1  + cos  6*)  =  4  cos2— 

197.  Examples  in  Quadrature  of  Plane  Curves.  The  quadra- 
ture of  a  figure  or  surface  is  the  finding  of  its  area.  For 
formulas,  see  §  66. 

1 .    Find  the  area  of  the  circle  x2+y2  =  r2. 
Here        area  =  4  j  (r2  —  x2)hdx 

r=7r^.   §185,  Ex.  8. 


-' 


x(r2-x2)h       r3  z[n-ix~ 
2  2  r 


EXAMPLES.  221 

For  the  segment  between  the  lines  x  =  a  and  x  =  6,  we  have 
area  =  2  J  (r2  —  cc2)  *  dx 


=  by/t-  —  b2-\-r  sin-1-  —a^/r2  —  a2—  i~a\\\  ' 
r  ?* 

2.    Find  the  area  of  one  branch  of  the  cycloid, 


x  =  r  vers-1-  —  V2  ry  —  y2. 

Area  =  2  C    y'dy       =  3***.  §  185,  Ex.  18. 

«^°    yj'lry—y1 

Hence  the  area  of  one  branch  is  three  times  that  of  the  gen- 
erating circle. 

3.    Find  the  area  of  the  tractrix. 


Here        ^  = y       .or dx=  -V ^ ~^-dy.  §195, Ex. 5. 

dx  Va2  -  y2  V 

"e  =jydx  =  ~  j  ^°2 - y2 c1,j  ; 

. ' .  area  =  —  4  I  Va2  —  y2  dy  =  ira2.  §  185,  Ex.  8. 

Hence  the  whole  area  enclosed  by  the  curve  is  equal  to  the 
area  of  a  circle  whose  radius  is  a. 

Xs 

4.  Find  the  whole  area  between  the  cissoid  y-  =  — ■ and 

....  2(6  —  x 

its  asvmptote. 

Ans.  Sttcc. 

5.  Find  the  area  between  the  lines  xry  =  a8,  x  =  b,  x  =  c,  and 

J  ~    '  Ans.  a3 - 

be 

6.  Find  the  area  of  both  loops  of  the  curve  a?y2  =  a262.r— Jrx'. 

Ans.  |a&. 

7.  Find  the  area  of  one  loop  of  the  curve  a2y*  =  x*(a2  —  x2) . 

Ans.  \a2. 


222  LENGTHS   AND   AREAS    OF   PLANE   CURVES. 

198.    To  find  the  area  of  a  curve  given  by  its  polar  equation. 

Let  p  be  any  point  on  the  curve  ab  referred  to  the  pole  oaud 

b  the  polar  axis  ox.     Take  od  =  1 ,   and  draw 

\Pf       the  arcs  db  and  pr  ;  then,  if  db  =  d6,  it  is  evi- 

x/K       dent  that  the  sector  opr  =  dA,  A  representing 

...'h  the  area  traced  by  the  radius  vector. 

Hence      dA  =  £pr  •  op  =  %p2d$  ; 


Fig.  55.  .'.  A  =  %  J  p2d$. 

Or,  let     dc  =  A0  ;  then  kp'=  Ap,  and  area  opp'=  A  A. 
Now        opk  <  opp'<  ohp',  or  £- A0  <  AA<^P  +  Ap^~  AO. 


2 


2        A0               2 

dA  _  p- 
'~dO~ll 

Examples. 

£,  or  A=%CfM0. 


1.  Find  the  area  of  the  first  spire  of  the  spiral  of  Archimedes, 
p  =  a6 ;  also  the  area  between  the  first  spire  and  the  second. 


Here        A  =  £  fa?6*d9  =  i  a203  =  \p26. 


O=0,  since,  if  the  area  be  estimated  from  the  pole,  A=0 
when  0  =  0. 

When  $=2tt,  p  =  r,  or  the  radius  of  the  measuring  circle ; 
and  A  =  %ttis.  Hence  the  area  of  the  first  spire  is  one-third 
of  the  area  of  the  measuring  circle. 

When  $  =  4 ir,  p  =  2r,  and  A  =  f  Trr2. 

But,  in  the  two  revolutions,  the  area  of  the  first  spire  has 
been  traced  twice ;  hence  the  area  between  the  first  spire  and 
the  second  is  \-n-r2  —  f-n-r2,  or  twice  the  area  of  the  measuring 
circle.  The  area  between  the  second  spire  and  the  third  is  four 
times  the  area  of  the  measuring  circle  ;  and  so  on. 

2.    Find  the  area  of  the  curve  p  =  a  sin  30. 
The  curve  consists  of  three  equal  loops  (Fig.  47).     Hence 
the  area  equals  three  times  the  area  of  the  first  loop. 

Ans.  \ira2. 


QUADRATURE  OF   SURFACES   OF   REVOLUTION.      223 

3.  Find  the  area  of  the  lemniscate  p2  =  a2  cos  2  6. 

The  integral  between  0  =  0  and  0  =  $tt  is  one-fourth  of  the 

whole  area.  A 

Ans.  a~. 

4.  Find  the  area  of  the  cardioid  p  =  a  (cos  0  -f- 1 ) . 

Ans.  %Tra2. 

199.   Examples  in  Quadrature  of  Surfaces  of  Revolution.   For 

formulas,  see  §  68. 

1 .  Find  the  area  of  the  surface  of  the  prolate  spheroid ;  that 
is,  of  the  surface  traced  by  the  revolution  of  the  ellipse 

y8  =  (l-e8)(o?-a?) 

about  the  axis  of  x. 

Area  =  2  fl ir ( 1  - e2) h (a2 - x2 ) hfl  +  (^Qfdx 

=  4  Tre  ( 1  -  e2)  I  f  Y4  -  ^Ycfce 

=  47re-  I    l  —  —x~)  dx 
tun  \e-         J 

.      b\~ i    fa2       »V      a2    .  _ieaflaeio-   ^     0 
=  47re-    £» (-  —  or    +  — sin  *—      §18o,  Ex.8. 
a\_      \er         J       2er  a  J0 

7rO"H sin  1e. 

e 

2.  Find  the  area  of  the  surface  of  the  prolate  spheroid  whose 
generatrix  is  Sy2  +  4  o^  =  36 . 

3.  Find  the  area  of  the  surface  generated  by  the  revolution 
of  the  cycloid  about  its  base. 

C2r  C2r  f       daj*\i 

Area  =2  I    2-n-yds=Air  I    yl\  +  -^\dy 

=  47rV2r|    y(2r  —  y)~^dy 

=  47rV27-[-f(4r+^)(2r-y)i]Jr     §179,  Ex.3. 


224 


CUBATUKE   OF   SOLIDS   OF   REVOLUTION. 


4.    Find  the  area,  of  the  surface  generated  by  the  catenary 
revolving  about  the  axis  of  x,  between  the  limits  0  and  b. 


Ids 


Here        S=2ir\  yds  =  Tra\    \ea-\-ea)t 

s=i7raJ0  W  +  e'jdx        §76,  Ex.  8. 

ra2(  vl      j±\      i 

=  tt    jVe"  —  e      J  +ab  \. 

5.    Find  the  area  of  the  surface  generated  by  the  revolution 
of  the  tractrix  about  the  axis  of  x.     (See  §  195,  Ex.  5.) 

Ans.  4xa2. 

200.   Examples  in  Cubature  of  Solids  of  Revolution.     The 

eubature  of  a  solid  is  the  finding  of  its  volume.     For  formulas 
see  §  69. 

1.    Find  the  volume  of  the  solid  generated  by  the  revolution 
of  the  cycloid  about  its  base. 

Here        dx  =  — ^  ^       :    .'.Try2dx  = — ^  ^ 


-\/2ry—  y2 


volume 


=  2tt  f 

c/0- 


fdy 


^/2ry  —  f 
5-n2^;        §185,  Ex.  19. 


•  V2  ry  —  y2 
that  is,  the  volume  is  five-eighths  of  the  circumscribed  cylinder. 

2.    Find  the  inclosed  volume  of  the  solid  generated  by  the 
revolution  of  the  parabola  y2  =2px  about  the  line  x  =  a. 

Let  mk  be  the  line  x=a  ;  let  b= ak[=  V2pa]  ; 
and  let  p  be  any  point  on  the  parabola ;  then 
oh  =  »,  and  ha  =  a—x.  Now,  if  bc  =  dy,  the 
volume  generated  by  the  revolution  of  bpdc 
about  mk  equals  dV; 

.-.  dV  =  Tr(a  —  x)2dy ; 


Fig.  56. 


volume  =  2tt  j    (  a—  ^—  j  dy  =  \%irba? 


3.    Find  the  volume  of  the  solid  generated  by  the  revolution 
of  the  cissoid  about  its  asymptote.  Ans.  2ir2a3. 


EXAMPLES.  225 

4.    Find  the  volume  of  the  solid  generated  by  the  revolution 
of  the  tractrix  about  the  axis  of  x. 

Since  y  =  a  when  x  =  0,  and  y  =  0  when  x  =  oo,  and 


dx  =  -V  -•''  dy; 

y 


y 

volume  =  2  I  Try2dx=  —  2tt  I  y  Va2  —  y-  dy  =  f -a3 

c/i=0  «/y=o 


201.  The  Calculus  is  often  of  great  aid  in  deducing  the  equa- 
tions of  curves.  The  equation  of  the  catenary  is  obtained  by 
its  use  in  §  76,  Ex.8. 

Examples. 

1.  Find  the  equation  of  the  tractrix. 

Here        ^  =  -        ^        •  §195,  Ex.  5. 

dx  (a-  —  y)l 

.  __    r(a2-y2)hc]y. 
J        y 

.-.  x=a\oga+(a2~y2)i-(a?-y2)l.      §  185,  Ex.  13. 
C=  0,  since  x=  0  when  y=  a. 

2.  Find  the  equation  of  the  curve  whose  subtangent  is  c. 

Here        y  —  [=subt.]  =  c; 
dy 

7       ^y 

.-.  dx=c  —  ; 

y 

.-.    x  =  —  logay+C. 
m 

If  c  =  m,  and  the  curve  pass  through  the  point  (0,  1),  we 

have  , 

»— log8y. 

3.  Find  the  equation  of  the  curve  whose  subnormal  is  c  tunes 
the  square  of  its  abscissa.  A  »      ->     i  ,  n 


CHAPTER  XVII. 
THE  METHOD  OF  INFINITESIMALS. 

202.  Infinitesimals  and  Infinites.  A  quantity  so  small  that  its 
value  cannot  be  expressed  in  terms  of  a  finite  unit,  is  said  to  be 
infinitely  small. 

An  Infinitesimal  is  an  infinitely  small  variable  whose  limit  is 
zero.  For  example,  if  y  =f(x),  Aa;  and  Ay  both  become  infini- 
tesimals as  Aa;  =  0.  Again,  any  variable,  when  near  its  limit, 
differs  from  its  limit  by  an  infinitesimal. 

"When  we  consider  several  related  infinitesimals,  we  choose 
arbitrarily  some  one  of  them  as  the  principal  infinitesimal,  and 
adopt  the  following  definitions  : 

Any  infinitesimal,  the  limit  of  whose  ratio  to  the  principal 
infinitesimal  is  finite,  is  an  infinitesimal  of  the  first  order. 

Any  infinitesimal,  the  limit  of  whose  ratio  to  the  square  of  the 
principal  infinitesimal  is  finite,  is  an  infinitesimal  of  the  second 
order. 

Any  infinitesimal,  the  limit  of  whose  ratio  to  the  nth  power 
of  the  principal  infinitesimal  is  finite,  is  an  infinitesimal  of  the 
nth  order. 

Hence,  if  i  represent  the  principal  infinitesimal,  vxi,  v2t2,  vsis, 
and  vntn  will  represent  respectively  an}-  infinitesimals  of  the  first, 
second,  third,  and  nth  orders,  in  which  vx,  v2,  v3,  and  vn  are 
variables  having  finite  limits,  from  which  they  differ  by  infini- 
tesimals. According  to  this  notation,  a  =  i,  a  =  v^i,  and  a  =  v2i2 
are  read  respectively,  "  a  =  the  principal  infinitesimal,"  "  a  =  an 
infinitesimal  of  the  first  order,"  and  "  a  =  an  infinitesimal  of  the 
second  order." 

A  quantity  so  large  that  its  value  cannot  be  expressed  in 
terms  of  a  finite  unit,  is  said  to  be  infinitely  large. 


GEOMETRIC   ILLTTSTBATION.  -I'll 

An  Infinite  is  an  infinitely  large  variable  that  increases  with- 
out limit.  Hence  the  reciprocals  of  infinitesimals  are  infinites, 
and  the  different  orders  of  infinites  may  be  represented  by  "V  '- 
w2t"2,  w3i-3,  etc.,  in  which  w15  tv2,  wai  etc.,  are  variables  having 
finite  limits. 

The  symbols  0  and  o>,  or -,  represent  respectively  absolute 

zero  and  absolute  infinity,  of  which  there  are  no  orders. 

203.  From  the  algebraic  symbols  for  infinitesimals  and  infi- 
nites of  different  orders,  the  following  principles  are  evident : 

1.  The  product  of  any  infinitesimal  and  an  infinite  of  the 
same  order  is  a  finite  quantity  ;  thus,  v2c  •  iv2r2  =  v2v:2. 

2.  The  order  of  the  product  of  two  or  more  infinitesimals  is 
the  sum  of  the  orders  of  the  factors  ;  thus,  rxi« r2r  =  r,'-.i\ 

3.  The  order  of  the  quotient  of  an}*  two  infinitesimals  is  the 
order  of  the  dividend  minus  the  order  of  the  divisor ;  thus, 

v2l2     v2 

4.  If  the  limit  of  the  ratio  of  one  infinitesimal  to  another  is 
zero,  the  former  is  of  a  higher  order  than  the  latter ;  thus, 

limit  —  =  limit— t  =  0. 

204.  Geometric  Illustration  of  Infinitesimals  of  Different 
Orders.  Let  cab  be  a  right  angle  inscribed  in  the  semicircle 
cab,  bd  a  tangent  at  b,  and  ae  a  perpendicular  to  m>.  From 
the  similar  triangles  cab,  bad,  and  aed, 
we  have 


and 


AD 

_ 

AB 

AB 

AC 

DE 

AB 



= 

— 

AD 

BC 

Fig.  57 


Suppose  a  to  approach  B  so  that  AB  =  i-      Since  limit  AB  =  '». 
and  limit  AC  =  en,  from  (1)  we  have 


228 


THE   METHOD   OF   INFINITESIMALS. 


limit  —  =  limit  — ■  =  0  ; 

AB  AC 

hence  ad  is  an  infinitesimal  of  a  higher  order  than  ab  (§  203,  4) . 
From  (2)  we  have 

,.      .,  DE        v      .,AB 

lmnt  —  =  lmnt  —  =  0  ; 

AD  BC 

hence  de  is  an  infinitesimal  of  a  higher  order  than  ad. 
Thus,  when  ab  =  i,  ad  =  v2<.2,  and  de  =  v3l3. 


Since        limit  1"^' 


Examples. 

1.  If  a  =  t,  of  what  order  is  sin  a? 
1,  sin  a  =  iv  (§202), 

and  limit  t\  =  1 . 

2.  If  a  =  t,  of  what  order  is  tana? 

3.  If  a  =  i,  of  what  order  is  1  —  cos  a? 

Since         ^  P~JSal  =  |,  1  -  cos  a  =  v2c  (§  202) , 

and  limit  v2  =  $■• 

4.  If  a  =  t,  show  that  sin  a— a=v3i3,  and  that  a  —  tana=vsiz. 

205.   First  Fundamental  Principle  of  Infinitesimals. 
Let  a  —  /?  =  c,  in  which  e  is  infinitely  small  in  comparison  with 
a  or  /3 ;  then 

-=l+i,  and  limit-J  =  0  ; 
fi  fi  fi 


.'.  limit-  =  limit 
fi 


l  +  £. 


fiA 


Hence,  if  the  difference  between  tivo  variables  is  infinitely  small 
in  comparison  icitk  either  of  them ,  the  limit  of  their  ratio  is  unity  ; 


EULE   FOR   DIFFERENTIATION.  229 

and,  by  §  135,  either  of  them  may  be  substituted  for  the  other  in 
any  problem  concerning  the  limit  of  the  ratio  of  two  variables. 

For  convenience  of  application,  this  principle  may  be  stated 
as  follows : 

In  problems  concerning  the  limit  of  the  ratio  of  two  variables, 

All  infinitesimals  of  the  higher  orders  may  be  dropped  from 
sums  of  infinitesimals  of  different  orders. 

All  infinitesimals  may  be  dropped  from  sums  of  finite  quantities 
and  infinitesimals. 

All  finite  quantities  maybe  dropped  from  sums  of  infinites  and 
finite  quantities. 


Cor.    If  a  =  /?+€,  limit-,   or  limit  \l  +  - 
p  P  L      fi. 


is   unity   only 


when  €  is  infinitely  small  in  comparison  with  /?. 

Hence,  conversely,  if  the  limit  of  the  ratio  of  two  infinitesimals 
is  unity,  their  difference  is  infinitely  small  in  comparison  with 
either. 

206.  Rule  for  Differentiation.  In  this  chapter  we  shall  regard 
the  increments  of  variables  as  infinitesimals.  Since  the  differ- 
ence between  a  variable  and  its  limit  is  an  infinitesimal,  and 
since 


limit 
Ax  =  0 


Ay 
Ax 


=/'(*) ; 


.-•^  =/'(»)  +  €,  or  Ay  =/'(» Az  +  cAz,  (1) 

in  which  e  is  an  infinitesimal. 

Now         dy=f(x)dx.  (2) 

The  value  of  dx  being  arbitrary,  for  convenience  we  shall,  in 
this  chapter,  suppose  it  to  be  equal  to  Ax. 

From  (1)  and  (2)  it  follows  that,  if  dx  =  Ax  =  i,  dy  and  Ay 
are  infinitesimals  whose  difference  is  infinitely  small  in  compari- 
son with  either,  and  therefore,  in  differentiating,  dy  may  be 
substituted  for  Ay  (§  205). 


230  THE  METHOD   OF   LNFESTITESIMALS. 

From  these  considerations  we  have  the  following  simple  rule 
for  differentiating  any  function  : 

Find  the  increment  of  the  function  in  terms  of  the  increments 
of  its  variables,  apply  the  principles  of  §  205,  and  in  the  terms 
remaining  replace  the  increments  by  differentials. 

Thus,  to  differentiate  a3,  let  y  =  Xs ;  then 

Ay  =  3  x2Ax  +  3  x(Ax)2  +  (Ax)\ 
Hence,  by  the  rule, 

dy  =  S  x2dx. 

Rem.  We  do  not  drop  the  infinitesimals  of  the  higher  orders, 
because  they  are  nothing,  or  comparatively  nothing,  when  added 
to  an  infinitesimal  of  the  first  order,  but  because  we  know  that 
they  do  not  appear  in  the  limit  of  the  ratio  sought.  Thus  the 
method  of  limits  is  the  basis  of  the  method  of  infinitesimals,  the 
difference  being  that  in  the  latter  we  use  infinitesimal  differen- 
tials, and  a  quantity  is  dropped  as  soon  as  it  appears,  when  it  is 
known  that  it  will  vanish  in  passing  to  the  limit  sought.  Any 
differential  equation  obtained  by  the  infinitesimal  method  must 
evidently  be  true  when  the  differentials  are  regarded  as  finite. 

Examples. 

1 .  Differentiate  u  =  xy. 

Here        Au  =  yAx  +  xAy  +  Ax  Ay ;  /.  du  =  ydx-\-xdy. 

2.  Differentiate  u  =  — 

y 

TT  A         yAx  —  xAy      .    ,        ydx  —  xdy 

Here        Au  = ;  .  .  du  =  ^ — *-• 

y2+yAy  y- 

3.  Differentiate  y  =  sin  a;. 

Here        Ay  =  sin  (x  -f  Ax)  —  sin  x 

=  cos  x  sin  Ax  —  ( 1  —  cos  Ax)  sin  x 
=  cos<cA:K-r-i>2i2; 


SECOND   FUNDAMENTAL   PRINCIPLE. 


231 


for,  when  Ax  =  i,  Ax  =  sin  Ax  +  nr  (§  48,  Cor.,  and  §  205,  Cor.), 
and  l-cosAaj  =  t'.,i2  (§  204,  Ex.  3). 

.'.dy  =  vo&xdx. 

4.    Find  the  differential  of  any  plane  curve. 
Let  As,  or  arc  pp'  in  Fig.  58,  be  i ;  then 

As  =  chord  pp'  +  va C-  §  48. 


=  VAor  -f  A?/2  +  v2l2  ; 
.'.ds  =  ^/dy~  -f-  dx*. 

5.  Find  the  differential  of  the  area  between  a  curve  and  the 
axis  of  x. 

Let  Ax ,  or  ae  in  Fig.  58,  be  t ;  then,  since  area  pdp'<  Ax- Ay, 
Az  =  area  abp'p  =  y Ax  +  v2i2 ; 
.'.dz  =  ydx. 

207.  Second  Fundamental  Principle  of  Infinitesimals.  Let 
a:,  a2,  a3,  •••,  an  be  any  infinitesimals  so  related  that,  as  n 
increases, 

limit  [ax  +  <*2  +  a3  H 4-  a>.]  =  c ; 

and  let  fii,  /?2,  y83,  •••  (3n  be  any  other  infinitesimals,  such  that 


in  which  e2,  <r2, 


a, 

are  infinitesimals. 


+  €„, 


(1) 


Clearing  equations  (1)  of  fractions,  adding,  etc.,  we  obtain 

A  +  &+-+fo-(a1  +  as  +...+an) 

=  a1€1+a2e2H (-<*,,*„• 

Let  8,  a  positive  infinitesimal,  be  greater  in  absolute  value 
than  any  of  the  infinitesimals  en  e2,  •••,  e„  ;  then  we  have  numeri- 
cally, 

(&+#,  +  - +&)-(ai  +  aa  +  ...+a.) 

<  8  (aj  +  a2  H f-  an) . 


232 


THE   METHOD    OF   INFINITESIMALS. 


But,  since  limit  8  =  0,  and  limit  (ax  +  a2  -\ h  «„)  =  c, 

limit[8(a1  +  a2H |-an)]  =  0. 

Whence  limit  [fii  +  fi2-\ \~  fin]  =  limit  [ax  +  a2  H +  an] . 

Hence,  (f  the  difference  between  two  infinitesimals  is  an  infini- 
tesimal of  a  higher  order,  either  may  be  substituted  for  the  other 
in  any  problem  concerning  the  limit  of  the  sum  of  infinitesimals, 
provided  this  limit  is  finite. 

n  T*     fi1  1       I  P2  1      I  Pn  1       I 

Cor.   If  —  =  1+£i,   —  =l  +  e2>  •••> —=l  +  «»» 

and  c^  +  c^-l — +an  =  c, 

then  limit  [ft  +  /?2  H h  ft]  =  a-i  +  «2  H \-an  =  c. 

208.   Integration  as  a  Summation.     Let  z  represent  the  area 
between  the  curve  opd  and  the  axis  of  x,  and  let  us  seek  the 
_  area  of  the  portion  obd.     At  the  several 

values  of  x,  as,  0,  oa,  oa',  and  oa",  let 
oa  =  aa'  =  a'a"  =  a'*B  =  Ax  =  dx ; 
then  the  corresponding  values  of  dz  are 
0,  avba',  a'p'b'a",  and  a"p"b"B,  while  those 
x  of  Az  are  opa,  app'a',  a'p'p"a",  and  a"p"DB, 
Fig.  59.  whose  sum  equals  obd.     Let  the  divisions 

oa,  aa\  etc.,  become  infinitesimals,  but  increase  in  number  so 
that  their  sum  will  continually  equal  ob  ;  then  Az  and  dz  both 
become  infinitesimals;  and,  since  opa,  pp'&,  etc.,  is  each  less 
than  Ace  •  Ay,  Az  —  dz  =  v2t2.  Therefore  the  limit  of  the  sum 
of  the  values  of  dz  is  equal  to  the  sum  of  the  values  of  Az 

(§  207,  Cor.).     Hence,  if  x'  =ob,  and  the  symbol    ^  Az  repre- 

sent  the  sum  of  the  values  of  Az  corresponding  to  the  different 
values  of  x  between  0  and  x',  we  have 

x=x'  x  =  x'  x=x' 

area  obd  =  £  Az  =  limit  £  dz  =  limit  £  ydx. 


v"  ^1 

b" 

p  jS 

b' 

p 

6 

But 


area  obd 


=  (ydx; 


■s 


ydx  =  limit  V  ydx. 

-r=0 


CENTRE   OF   GRAVITY.  233 

Hence,  when  differentials  are  infinitesimals,  integration  may 
be  viewed  as  the  summation  of  an  infinite  series  of  infinitesimals. 

209.  Centre  of  Gravity.  The  centre  of  gravity  of  a  body  is  a 
point  so  situated  that,  if  it  be  supported,  the  body  will  remain  at 
rest  in  whatever  position  it  niay  be  placed.  An  element  of  any 
quantity  or  magnitude  is  an  infinitely  small  portion  of  it.  The 
product  of  the  weight  of  a  body  by  the  distance  of  its  centre  of 
gravity  from  a  given  plane  is  called  the  moment  of  the  body  with 
respect  to  that  plane.  The  moment  of  a  bod}*  is  the  sum  of  the 
moments  of  its  elements.  Hence  the  distance  of  the  centre  of 
gravity  of  a  body  from  a  given  plane  equals  the  sum  of  the 
moments  of  its  elements  divided  by  the  weight  of  the  body. 
The  bodies  here  considered  are  supposed  to  be  of  uniform  den- 
sity ;  hence  their  weights  are  proportional  to  their  volumes. 

The  advantage  sometimes  gained  by  viewing  integration  as  a 
summation  is  illustrated  in  deducing  formulas  for  finding  the 
centre  of  gravity. 

210.  To  find  the  centre  of  gravity  of  any  plane  surface.     Let 
(cc,  y)  be  any  point  on  the  curve  OP?t  referred  to  the  axes  ox 
and  oy,  and  let  x0  and  y0  represent  respec- 
tiveby  the  distances  of  the  centre  of  gravity  of 
any  portion  of  the  surface  xow  from  the  planes 
oy  and  ox,  which  planes  are  perpendicular  to 
that  of  the  figure.      The  differential  of  the 
area  xon  is  ydx  ;  now,  if  dx  =  ab  =  i,  ydx  will  a        b 
differ  from  abp'p,  the  corresponding  element 

of  this  area,  by  v.2c  ;  and  the  distance  of  the  centre  of  gravity 
of  this  element  from  the  plane  oy  will  differ  from  x  by  i^i; 
hence  xydx  will  differ  from  the  moment  of  this  element  with 
respect  to  the  plane  oy,  by  v2i2.  Therefore,  between  x  =  a  and 
x  =  6,  the  sum  of  the  moments  of  the  elements 

=  limit  ;£  xydx  =   1  xydx ; 

x=a  K/a 

Xb                       fb 
xydx       I  xydx 
=  *_ . 

"    °        area        '    f"  . 

Xy 


234  THE   METHOD   OF   INFINITESIMALS. 

Again,  the  centre  of  gravity  of  ydx  is  evidently  \y  from  the 
plane  ox  ;  hence  \y2dx  will  differ  from  the  moment  of  the  cor- 
responding element  with  respect  to  the  plane  ox,  by  v2c2.  Hence, 
between  x  =  a  and  x=  &,  the  sum  of  the  moments  of  the  ele- 

y2dx ; 


Xb  S*b 

y2dx         I  y2dx 

a  ~*~rT 

I  ydx 


■Vo  =  i 


If  the  curve  be  symmetrical  with  respect  to  ox,  x0  is  evidently 
the  same  for  the  whole  area  as  for  the  half,  and  y0  is  zero. 

211.  To  find  the  centre  of  gravity  of  any  plane  curve.  Let 
(xoi  Vo)  De  the  centre  of  gravity  of  any  arc  of  a  plane  curve 
whose  length  is  represented  by  s.  Now,  when  ds  =  i,  xds  dif- 
fers from  the  moment  of  the  corresponding  element  of  the  curve, 
with  respect  to  the  plane  oy  (Fig.  GO),  by  v2t2.  Hence,  between 
x  =  a  and  x  =  b,  the  sum  of  the  moments  of  the  elements 


=  I  xds, 


and  (xds 

v  —  ^a 
s 

In  like  manner,  we  obtain 

I  yds 

212.  To  find  the  centre  of  gravity  of  a  solid  of  revolution. 
The  differential  of  a  solid  of  revolution  whose  axis  is  the  axis 
of  x,  is  -n-yhlx ;  hence,  if  dx  =  t,  irxifdx  will  differ  from  the 
moment  of  the  corresponding  element  of  the  solid,  with  respect 
to  the  plane  oy,  by  v2t2 ;  and  therefore,  between  x  =  a  and  x=b, 
the  sum  of  the  moments  of  the  elements 


=  I  Trxy2dx. 


EXAMPLES. 


235 


7r  I  xy-dx       1  xy-dx 
•   volume  "  pfdx 


As  the  centre  of  gravity  must  evidently  be  on  the  axis  of  revo- 
lution, the  formula  given  above  entirely  determines  it. 

Examples. 

1.    Determine  the  centre  of  gravity  of  a  circular  arc  bad. 

Let  the  extremity  d  be  (x' ,  y') . 

Here         y2=  2  rx  —  xr;  yd 

7         (r—x)dx 
.-.dy=  \        >\ 

V2r*  —  xr 

rdx 


,:  ds  =  V<t*r  +  dy-  = 


I  it*(Zs 


■\/2rx—xr 


/■// 


TT                                    ry'     r  chord  bd 
Hence      ce  =  r  —  ae  =  —  = 


2.    Find  the  centre  of  gravity  of  a  segment  of  a  circle. 
Using  the  equation  of  the  circle  referred  to  its  centre,  we  have 


I  xydx        I  (r2—  x2)hx 


dx 


_^0-2-'r)*-:H>-2-&2)* 
area 

lfa  =  0,  and&  =  r,  then  area  =  ^7rr,  and  we  have  xa  = 
when  the  segment  is  a  semicircle. 


3tt 


236  THE   METHOD   OE   INFINITESIMALS. 


3.    Find  the  centre  of  gravity  of  a  parabolic  area. 


Ans.  x0  =  ^x'. 


4.  Find  the  centre  of  gravity  of  a  right  cone. 
Here  y  =  ax,  and  volume  =  -kiry2x ; 

■k  j  xyhlx      7r  I  a^dx 

volume  ^7raV3 

that  is,  the  distance  of  the  centre  of  gravity  from  the  vertex  is 
three-fourths  of  the  axis. 

5.  Find  the  centre  of  gravity  of  a  segment  of   a  prolate 
spheroid.  —  (a  ob" -*-£»") 

Ans.  x0  — - 

volume 

"When  x'=  a,  x0  =  %a. 

j  xyds 

6.  Prove  that  a0=^ is  the  formula  for  finding  the  cen- 

(yds 

i/a 

tre  of  gravity  of  an}-  surface  of  revolution. 


Press  cf  Berwick  &  Smith,  it8  Purchase  Street. 


I^p* 


LD 


Uoiversgrkeley 


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GENERAL  LIBRARY  -  U.C.  BERKELEY 


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